CreationDate
stringlengths 23
23
| Body
stringlengths 57
10.9k
| Tags
stringlengths 5
114
| Answer
stringlengths 39
16.3k
| Id
stringlengths 1
5
| Title
stringlengths 15
149
|
|---|---|---|---|---|---|
2016-10-19T19:35:50.790
|
<p>I am currently trying to attach a small flat piece of plastic (no bigger than 2", could be smaller) to a 1.9" hollow aluminum tube bent into a 2 foot radius torus, spinning at 2Hz.
My first and normal approach would be to drill a tapped hole in the hollow tube; however this is not an option in this situation, as the inside of the cylindrical tube is kept in a vacuum, and has to be in order for the project we are working on to work.
My second approach would be to mill the plastic to match the diameter of the tube, but the problem remains; how would I attach the resulting plastic to the tube?</p>
<p>I have access to a machine shop, so I can do some basic mill and lathe stuff in order to accomplish this.
How should I do this?</p>
|
|mechanical-engineering|design|
|
<p>One way would be to clamp the outside of the pipe with a two-piece shaft collar. Properly tightened, a shaft collar will resist pretty heavy loads before it slips. You could tap the shaft collar and attach your plastic part as you normally would before attaching the collar to the pipe under vacuum. Note that a shaft collar will compress the pipe, so there is some risk of crushing the pipe, especially since it is already under inwards pressure. I doubt this will be a practical concern, but it is worth doing at least a rough check if you are using a thinner pipe wall.</p>
<p>One problem might be that shaft collars are typically manufactured for shaft sizes not pipe sizes, so you probably won't be able to find a perfectly sized collar. Since you have access to a shop, you could machine your own from scratch just for the pipe, or you could buy an 1 7/8" collar (like McMaster Carr 6436K26) and machine the ID slightly larger. In a pinch you could open up the ID with a grinder too, but turning or milling would be a much better option.</p>
|
11928
|
Ways to attach stuff to a cylindrical tube that is in vacuum?
|
2016-10-21T06:59:42.180
|
<p>I'm was looking into the <code>ASME B16.5</code>Pipe flanges and pipe fittings and it has 2 dimension tables metric and inches.</p>
<p>What I found was rather strange for me. I was looking at a 2 inch welding neck flange, and the hub diameter must match the pipe size.</p>
<p>For the metric table this is <code>60.3mm</code> and it is the same in all pipe sizes charts that I have available. When I look at this dimension in the Inch table, it is <code>2.38in</code> when I convert this to <code>mm</code> this varies from the metric table..
<code>2.38in * 25,4 = 60,452mm</code></p>
<p>So there is a <code>,152</code> difference between these values. <strong>What is correct?</strong> These differences are to be found on more flanges then just the 2" flange size. </p>
|
|process-engineering|piping|pipelines|
|
<p>As of the 2013 edition of ASME B16.5, the scope section explains:</p>
<blockquote>
<p>This Standard states values in both SI (Metric) and
U.S. Customary units. As an exception, diameter of bolts
and flange bolt holes are expressed in inch units only.
These systems of units are to be regarded separately as
standard. Within the text, the U.S. Customary units are
shown in parentheses or in separate tables that appear
in Mandatory Appendix II. The values stated in each
system are not exact equivalents; therefore, it is required
that each system of units be used independently of the
other. Except for diameter of bolts and flange bolt holes,
combining values from the two systems constitutes nonconformance
with the Standard.</p>
</blockquote>
<p>This is actually a pretty common practice where metric and inch systems are regarded separately. While the numbers may not match exactly, the intention is that they work as a system within each unit of measure. Of course, as far as I'm aware, you would not buy a metric flange for an NPS or DN size pipe separately, they will be the same product manufactured to some company's internal drawings, so in essence the standard is accepting that any rounding errors are within tolerances intended by the standard. One reason they don't allow intermixing when interpreting the standard is that they don't want you to be able to cherry-pick the most permissive set of dimensions and tolerances for each feature.</p>
<p>The forward also notes:</p>
<blockquote>
<p>The 2003 Edition included metric units as the primary reference units while maintaining U.S.
Customary units in either parenthetical or separate forms.</p>
</blockquote>
<p>So it would be reasonable to consider the metric information in the standard to be slightly more authoritative than the inch information.</p>
<p>The forward to ASME B36.10M-2004, which covers the dimensions of the actual pipe, explains</p>
<blockquote>
<p>The standard was revised in 1978 to include SI metric dimensions. The outside diameter and
wall thicknesses were converted to millimeters by multiplying the inch dimensions by 25.4.
Outside diameters larger than 16 in. were rounded to the nearest millimeter, and outside diameters
16 in. and smaller were rounded to the nearest 0.1 mm. Wall thicknesses were rounded to the
nearest 0.01 mm. These converted and rounded SI metric dimensions were added to Table 2. A
formula to calculate the SI metric plain end mass, in kilograms per meter, using SI metric diameters
and thicknesses was added to section 5. The SI metric plain end mass was calculated and was
added to Table 2. These changes in the standard were approved and it was designated an American
National Standard on July 18, 1979.</p>
</blockquote>
<p>So they do acknowledge some use of rounding. For 2" NPS pipe they list the OD as 2.375 inches and 60.3mm (DN 50 size.) There there is only an error of .025mm which is consistent with their rounding philosophy. While your observed error of .152mm is definitely larger than I can directly justify, it is unlikely to cause a serviceability issue in a welded joint - .006" in below the manufacturing tolerances of most metal parts that are to be welded. Tolerances that small are usually achieved by precision machining. I would not be concerned about the actual function of the joint. </p>
<p>To address your specific question in the title, the pipe should be sized according to its own spec. That standard may have its own dimension tables, or it may reference ASME B36.10M and then specify manufacturing tolerances on top of it. In the US, if you'r ejust looking at plain black pipe, the relevant standard would be ASTM A36, but a multitude of standards exist for other materials, tighter tolerances, and application-specific pipes.</p>
|
11947
|
What is the correct pipe size for a 2inch pipe?
|
2016-10-21T17:21:51.757
|
<p>Is there something similar to the Section Modulus in the Flexure Formula</p>
<p>$$\rho_{max} = \dfrac{M_{max}}{S}$$</p>
<p>where $S = \dfrac{I}{c}$ for shear stress formula?</p>
<p>So in</p>
<p>$$\tau_{max} = \dfrac{VQ}{IT}$$</p>
<p>then there should be a Shear Section Modulus published which would be $Q(\text{of centroid}) / IT$</p>
<p>So you can get $\tau_{max} = V * \text{Shear Section Modulus}$.</p>
|
|stresses|moments|shear|
|
<p>What you're looking for is commonly referred to as the <strong>shear area</strong> of the section (see <a href="https://knowledge.autodesk.com/support/robot-structural-analysis-products/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/Robot/files/GUID-50986085-D4AB-41C5-9B21-CF9619B8EB8B-htm.html" rel="nofollow">here</a> for a list of shear areas).</p>
<p>Consider a solid rectangle with depth 20mm and width 10mm. From <a href="https://skyciv.com/free-moment-of-inertia-calculator" rel="nofollow">this calculator</a> we can see that the section properties are:</p>
<p>$I = 6667 \text{ mm}^4$</p>
<p>$Q = 500 \text{ mm}^3$</p>
<p>$t = 10 \text{ mm}$</p>
<p>So the shear area is:</p>
<p>$A_{s} =\dfrac{It}{Q}=\dfrac{(6667)(10)}{500} = 133.3 \text{ mm}^2$</p>
<p>Therefore the shear stress is:</p>
<p>$\tau_{max}=\dfrac{VQ}{It}=\dfrac{V}{A_s} = \dfrac{V}{133.3 \text{ mm}^2}$</p>
<p>Compare this to the <a href="https://knowledge.autodesk.com/support/robot-structural-analysis-products/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/Robot/files/GUID-50986085-D4AB-41C5-9B21-CF9619B8EB8B-htm.html" rel="nofollow">resource</a> I shared earlier and you can see that for a thick walled rectangular section the shear area (denoted by W in the resource) is:</p>
<p>$A_s=\dfrac{2}{3}hb = \dfrac{2}{3}(20)(10) = 133.3 \text{ mm}^2$</p>
|
11953
|
Shear Stress / Flexure Stress Formulas / Tables
|
2016-10-22T07:03:41.303
|
<p>I am trying to solve the following problem:</p>
<blockquote>
<p>A masonry chimney having the shape of a conical frustrum is 25 m high.
The external diameter at the top and internal diameter at the bottom
is 2 m. The chimney is 0.5 m thick at its base. If the weight of the
chimney is 1800 kN, find the uniform horizontal wind pressure that may
act on the chimney per unit projected area of the chimney in order for
tension at the base to be just avoided.</p>
</blockquote>
<p>I am using the equation:</p>
<p>$$\mathrm{Centroid} (\bar{y}) = \dfrac {A_{1}y_{1} + A_2y_{2}} {A_{1} + A_{2}}$$</p>
<p>where $A1$ and $A2$ are the shaded areas in the figure below. Observe that the object is a chimney, so it is hollow in the centre portion; we are not adding up the area of that.</p>
<p><a href="https://i.stack.imgur.com/tisSD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tisSD.png" alt="Cross-sectional diagram of the chimney with the area on the left marked A1 and the area on the right marked A2"></a></p>
<p>The centroid becomes:</p>
<p>$$ \bar{y} = \dfrac{ (\dfrac{1}2 *0.5 *25) * \dfrac{25}3 + (\dfrac{1}2 *0.5 *25) * \dfrac{25}3}{(\dfrac{1}2 *0.5 *25) +(\dfrac{1}2 *0.5 *25) }$$</p>
<p>$$\bar{y} = 8.333$$</p>
<hr>
<p>However, I created a 3D model of the solid chimney in CAD and found the centroid to be 8.035, which does not match my calculated centroid. It is close to 8.333 but not exactly the same as my $\bar{y}$.</p>
<p>When I calculate the total wind pressure, which centroid should I use? The one I calculated by hand from the 2D cross-section, or the one I generated in CAD from the 3D model?</p>
<p>The CAD generated centroid is for a 3D solid and $\bar{y}$ calculated is for that central section.</p>
|
|structural-engineering|geometry|
|
<p>I believe that you are confusing yourself by introducing the 3D aspect when building your CAD model, and yet trying to determine the answer by hand using a 2D calculation.
Due to the volume being different to an area, this will skew your answer.</p>
<p>I hope this helps nudge you in the right direction</p>
|
11958
|
How should I calculate the centroid of a hollow structure when analyzing a wind load?
|
2016-10-22T23:55:50.420
|
<p>Cavitation and boiling are names for phenomenon that both involve sudden appearance of bubbles of vapor within a liquid, and in both cases they happen when the local hydrostatic pressure is lower than the vapor pressure of the fluid, but that doesn't necessarily mean they are the same thing.</p>
<p>In <a href="https://youtu.be/Lwk9Bi3j58o?t=105" rel="nofollow noreferrer">this video of an electrical heating element in water</a>, between 01:00 and 02:00 the sound produced by rapid bubble collapse gets louder and louder but there are few visible bubbles. Is the process that produces this sound considered <em>boiling</em> or <em>cavitation</em>? What is the distinction?</p>
<p>I've left a <a href="https://space.stackexchange.com/a/18573/12102">provisional answer</a> to the related question on another SE site: <a href="https://space.stackexchange.com/q/18486/12102">How (actually) do sub-cooled propellants reduce cavitation within turbo pumps and make feed easier?</a> I couldn't bring myself accept <a href="https://space.stackexchange.com/a/18496/12102">the answer</a> on that question that starts with the claim, "cavitation is boiling."</p>
<p>Although they are related, <em>what fundamentally distinguishes cavitation and boiling as different phenomena?</em></p>
|
|fluid-mechanics|thermodynamics|
|
<p>From a corrosion science standpoint, cavitation refers to what happens when a vapor bubble collapses <em>against the surface upon which it was formed</em> and thereby damages it. </p>
<p>The vapor bubble can be formed in the separated flow or "vacuum pocket" behind an overspeeding propeller blade or as a local vapor explosion on the surface of a hot object. </p>
<p>In the case of a hot object, the bubble collapse can occur either because the top of the vapor bubble moves far enough off the hot surface to encounter cooler fluid, which triggers its condensation and collapse (this is exactly what makes the kettle or heating element "roar" right before the bulk of the water gets hot enough that the whole of it begins to boil) or if the heat source is abruptly shut off immediately after the vapor explosion occurs. In either of these two cases, the bubble very quickly shrinks down to a point and the inrush of water behind it then strikes that point with pressures of order ~hundreds of atmospheres, on areas of order ~1 to 10 square microns. </p>
<p>The accumulation of cavitation damage is the leading cause of wearout mode failures in the heating elements of a thermal inkjet printhead, in which the cavitation is vigorous enough to beat holes down through the tantalum metal and/or silicon nitride or carbide which form a protective layer on top of the heaters. </p>
|
11963
|
What fundamentally distinguishes cavitation and boiling as different phenomena?
|
2016-10-23T22:21:47.787
|
<p>I'd like to understand the new SRI "Abacus" transmission, recently reported at
<a href="http://spectrum.ieee.org/automaton/robotics/robotics-hardware/sri-demonstrates-abacus-rotary-transmission" rel="nofollow">http://spectrum.ieee.org/automaton/robotics/robotics-hardware/sri-demonstrates-abacus-rotary-transmission</a></p>
<p>I haven't been able to find any information on it that doesn't seem derivative of the IEEE story.</p>
<p>I'm curious about how one determines the reduction ratio, and what the "grooves" look like mathematically (for example, the function from angle to width).</p>
<p>Judging from the pictures, this one seems to have
6 setscrews on the inner race,
7 "beads", and
8 setscrews on the outer race.</p>
<p>It's hard to tell the number of cycles for the grooves, but the ones in the outer race seem to be aligned with the setscrews.</p>
<p>They say the beads can have any profile you like; it seems that must then determine the groove profiles, but how would you get from one to the other?</p>
<p>I've tried to reason out the relationship between number of groove cycles on each race, diameter profile of the beads, offset of the driveshaft's attachment to the inner cam, and the resulting reduction ratio. But my intuition is failing even after watching the video many times.</p>
<p>Can anyone explain the relationships involved here?</p>
<p>Thanks!!!</p>
|
|mechanical-engineering|gears|bearings|mechanisms|
|
<p>It's just a fancy rolling cycloidal drive (not that that being simple makes it any less cool). It really bothers me that all the available articles only compare it to harmonic drives when it only bears a passing resemblance to them. Probably because it's owned by Harmonic Drive LLC. The bearing races can be abstracted as wire ring races that have been bent sinusoidally. The reason the metal one has a conical bearing is so that the races are easier to manufacture, since they're just sinusoidal rather than cycloidal, like they'd almost certainly have to be for the round bearing, and good luck milling that. What also bothers me about the video posted about it is that the guy claims it is a 1:3 drive. Well, the formula for the gear ratio of a cycloidal drive is P-L/L where P is the number of outer ring pins, in this case the number of periods of the outer sinusoidal wave, and L is the number of lobes on the cycloidal disk, in this case the number of bearings. (I think you're right about the set screws) (8-7)/7 = 1/7, not 1/3. </p>
|
11976
|
How does the SRI "Abacus" rotary transmission work?
|
2016-10-24T00:45:02.897
|
<p>I'm stuck with how to work out the total pressure and centre of pressure in this example.</p>
<p>At an outdoor boundary between two properties there is a stepped change in ground level of the clay soil. The ground abruptly changes level by H (roughly 0.9 - 1.2m if it matters). A vertical retaining wall is present, whose design is not relevant here. The soil's depth is much greater than the wall height so there are no relevant local boundaries or property changes in the soil below the retaining wall (however deeply it is embedded in the soil). There are no buildings or other sources of pressure on the soil other than its own weight and soil mechanics, and hydrostatic effects of any water it contains. There is considerable rainfall at times, so the soil will periodically become saturated. </p>
<p>I'm ignoring for now, any other issues, such as bearing, shear and other failure modes, but for safety I should probably assume active rather than passive pressure, I think. I want to find the maximum design lateral pressure such a wall must resist in the presence or absence of built-up hydrostatic pressure, and the height above the lower ground level at which that pressure should be taken as being exerted for calculation purposes. </p>
<p>Specifically I think what I'm trying to find is the maximum design lateral pressure exerted by the soil and any ground-water on the wall, and the point at which that pressure is exerted for calculation purposes (a) if there are ample weepholes/drainage, (b) if there are no weepholes/drainage. I'm expecting the answer to depend on the characteristics of the specific clay to an extent, but I can't work out which characteristics are critical and how much they affect the answer.</p>
<p><strong>Update</strong> Sketch as requested:</p>
<p><a href="https://i.stack.imgur.com/ZSc0P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZSc0P.png" alt="Sketch"></a></p>
|
|structural-engineering|civil-engineering|soil|retaining-wall|
|
<p>Usually since hydrostatic pressure along with freezing and thawing are such a pain (quite hard to figure out the pressure from freezing water in soil), in practice I try and keep my retaining walls 100% water free.</p>
<p><a href="https://i.stack.imgur.com/04CxQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/04CxQ.png" alt="Typical retaining wall construction"></a></p>
<p>However, if you were to design the wall for lateral earth pressure of clay + water table pressure, I found in my old notes some active, rest and passive values for clay which are around :</p>
<p>$k_a = 0.6 \\ k_o = 0.7 \\ k_p = 1.7$</p>
<p>Usually you will start with a hypothesis (which side is active or passive or at rest), and then confirm that your had chose correctly with maximum displacements. Since your aren't analysing the wall yet, we will keep that for later.</p>
<p>For safety, you should probably at least consider at rest pressure on both side (use $k_o$ for both sides) and verify your wall displacements afterwards (If you consider active pressure on the right hand side, this means your wall is relatively flexible and the soil can mobilise some of its shear stiffness).</p>
<p>Usually for safety, we also put the water table all the way up to the level of the ground, since this can easily happen after a storm in non-draining clay. You then apply your basic soil equations, first dry-soil pressure and add water pressure (both triangular loads). In this case we assume there is no external load on top of the right hand side of the wall.</p>
<p><a href="https://i.stack.imgur.com/h4Bnf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h4Bnf.png" alt="Retaining wall pressures"></a></p>
<p>The equations for the maximal pressure are the same for each side:</p>
<p>$ q_{clay} = k * \gamma_{clay} * H \\ q_{water} = k * \gamma_{water} * H $</p>
<p>If ever you decide to design this wall, you will see that your critical moment and shear will be at the base, even if you have some soil supporting on the left hand side. If you have enough movement to mobilise active pressure on the right and passive pressure on the left, then you can simply use the same equations with $k_a$ on the right and $k_p$ on the left.</p>
|
11979
|
Total lateral pressure from clay soil when hydrostatic pressure is also present
|
2016-10-24T04:09:06.450
|
<p>I'd like to know how do the authors of Mechanical Engineering books(books of engineering dynamics, mechanics of materials, fluid mechanics...) and some university professors of Mechanical Engineering formulate those beautiful problems and exercises ? Especially those involving real situations. That is, where do they get those problem situations ? Almost always involving a real mechanical system, thermal system, fluid system...</p>
|
|mechanical-engineering|education|
|
<p>They look them up in books, articles, standards, catalogues or they observe their surroundings. Mechanical engineering has been around for a while so many of the problems are known; Looking up a existing solution is quite feasible. </p>
<p>For example back when i was teaching mechanism design, many of my exercises were from things that one could be observed at the premises of the building itself. The benefit of this was that students could actually go and observe for themselves. Door closers, cupboard hinges even the cafeteria washing machine hood were all pretty good sources of exercises.</p>
<p>Catalogs are also great as they usually include drawings for the things. Ikea for example has extremely well done vector isometric pictures of everything they sell. Same applies to industrial vendors such as robot manufacturers and heavy construction gear producers. Who often even give out full 3D models of their outer geometry for design purposes (which allows you to repose and build your own highly accurate images). Once you have an idea for a source and some vendor names you can usually find something relevant pretty quickly.</p>
<p><a href="https://i.stack.imgur.com/zY5k2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zY5k2.png" alt="enter image description here"></a></p>
<p><strong>Image 1</strong>: A image of Kuka KR6 SIXX and a overall instructions on how image was made plus model download can be found <a href="https://graphicdesign.stackexchange.com/questions/59032/from-a-cad-model-to-an-illustration-drawing-how-to/59071#59071">here</a>. </p>
<p>Articles are great as they can include all the relevant math needed, although they tend to be very niche stuff and may or may not have good images. A usually better source is some of the more comprehensive mechanics books that exist (like <a href="https://accessengineeringlibrary.com/browse/mechanisms-and-mechanical-devices-sourcebook-fifth-edition#Preface01" rel="nofollow noreferrer">this one</a>). These books are wast catalogs of interesting mechanisms and have information about relevant design problems related to them.</p>
<p>Lastly, drawing images with CAD/hand is, or at least used to be part of the education of mechanical engineers. So taking what you have seen and sketching it up should not be a overwhelming task. But then I am also an illustrator so i tend to draw stuff just to keep my skills up.</p>
|
11980
|
Formulation of problems and exercises in Mechanical Engineering
|
2016-10-24T13:04:02.103
|
<p>I have a Li-Po battery charger that comes with its own cooling fan which, like most of the cooling fans, blows hot air out of the charger to cool the charger down.</p>
<p>I have a spare fan and I was wondering whether it should blow ambient air towards the charger or draw the hot air away from it?</p>
<p>It will be an external fan outside the casing of the charger. Charger itself is roughly the size of a normal Wi-Fi Adapter and the auxiliary fan has ben taken from a computer casing.</p>
<p>I know it does not really make much of a difference but was just curious as to which setup will perform better in order to dissipate the heat generated a little better.</p>
<p>Supplementary question:</p>
<p>If I had 2 extra fans would it be better if one was blowing air towards the charger and other was drawing away or both should be doing the same thing?</p>
|
|thermodynamics|cooling|
|
<p>The air coming out of the fan will be highly turbulent, wheras the air being sucked into the fan is less so. Hence blowing at the target will increase heat transfer. On the other hand, if you are looking at a casing, a fan blowing into the case will also blow dust into the case. A fan blowing out of the case will typically suck in less dust, as the fluid velocity at the opening of the case will typically be lower than directly next to the fan.</p>
<p>Finally, the choice of two fans in serial (one blowing, one sucking) or parallel (both blowing or sucking) depends on what you are trying to achieve (i.e. the <a href="http://www.greenheck.com/library/articles/42" rel="nofollow">system curve</a> of the device to be cooled). Two fans in series will increase the total pressure you can build up, and is a good idea if you need to overcome a high pressure loss. Two fans in parallel will increase your flow rate if the pressure loss of your system is relatively low.</p>
|
11992
|
Should an auxiliary cooling fan blow air or draw it?
|
2016-10-25T22:05:20.053
|
<p>A certain truck has 5 (per side) *2 sides= 10 bolts holding the body to the frame. The bolts are marked 10.9 and the area of the shank is 153 sq mm. From my math I got 3e8 lbs for all 10 bolts together.</p>
<p>From a mechanical engineering point of view, how would you describe the mechanical characteristics of these bolts?</p>
<hr />
<p>Edit:</p>
<p>Thank you for the discussion. I was dividing by the bolt section area; I should have been multiplying. Now the answers make much more sense.</p>
<p><strong>Each bolt has a tensile strength of about 32,000 lbs * 10 bolts = 320k lbs.</strong></p>
<p>Of course the more interesting tests in a vehicle application are the shear loads you would encounter in a collision, but that is also a MUCH more complicated question.</p>
|
|mechanical-engineering|
|
<p>The nominal capacity of a high tensile bolt are defined by its designation eg 10.9 here the first figure (10) is the ultimate tensile stress in hundreds of MegaPascals and the second figure (after the point) is the yield stress as a fraction of the UTS ie 9 represents 9 tenths or 90% </p>
<p>Conveniently if you multiply the yield stress (MPa) by the bolt section (mm) you get an answer for the yield load in tension of the bolt in Newtons. </p>
<p>Having said that the static tensile strength of the bolt is only part of the capacity of a bolted joint and you need to consider any loads in other direction as well as the effect of dynamic loading. </p>
<p>Also the preload on a bolt will have a significant effect on how it behaves, not all joints are preloaded but when they are external shear loads are carried by friction between the mating surfaces. It is also possible for bolted joints to fail in ways other than the bolt breaking eg by pulling through the member they are attached to. </p>
<p>Equally the design load for a structure may incorporate a factor of safety which puts the rated safe working load at some fraction of the theoretical maximum and where you have multiple bolts you can't always assume that they are equally loaded. </p>
<p>This is further complicated in something like a vehicle chassis in that the members are also carrying other loads as well.</p>
<p>Finally I would add that if you are working from metric units your answer should be metric as well, convert your final answer <em>as well</em> if necessary but having the problem and the answer in mixed units makes errors more likely and more difficult to identify. </p>
|
12021
|
How much do these bolts hold? Marked 10.9, with shank area of 153 sq mm
|
2016-10-26T05:59:40.567
|
<p>I have an engineering application where I would like to use a single 24v DC Stepping Motor to power two different axles of opposing direction. Basically when the motor turns clockwise, Axel A will spin and Axel B will be stationary. When the motor turns counter-clockwise, Axel A will be stationary and Axel B will spin.</p>
<p>I need it to be as cheap and simple as possible. So far, the best solution that I could come up with to accomplish this is by using two gears similar to the bicycle Freewheel / Freehub. The problem is, I cannot for the life of me find a gear that is similar to that application, but for uses outside of bicycles (possibly because I don't know that specific gear's name). Can anyone recommend a gear or something that is similar to the application that I need?</p>
<p>I have a general idea of the function that I need, but I don't know the name of this type of ratcheting gear.</p>
|
|mechanical-engineering|gears|
|
<p>You can get <a href="http://www.bocabearings.com/bearing-types/one-way-bearings" rel="nofollow">one way bearings</a> which rotate freely on one direction and lock in the other.</p>
<p>So if you use one to attach a gear or pulley to your motor output shaft then the pulley will be coupled to the shaft in one direction and not in the other so with two mounted in opposite orientation each with a pulley driving a separate shaft (A and B) you should be able to achieve what you want.</p>
|
12025
|
Looking for a gear similar to bicycle freewheel
|
2016-10-26T13:31:09.983
|
<p>I attend to engineering design of a new irrigation system, comprising 4 pump stations and 3 lengthy water mains between them.</p>
<p>Definitions by ISO 12006-2:</p>
<p>Construction Entity is an independent construction result of significant scale serving at least one user activity or function.</p>
<p>Construction Complex means two or more adjacent construction entities collectively serving one or more user activity or function.</p>
<p>In my case, the irrigation system must be identified as a construction complex, comprising 4 pump stations + 3 pipe mains = 7 'construction entities'.</p>
<p>How can I collectively and separately name the mentioned parts of the project? (In my native language those all are 'objects', however, definition for 'construction object' by ISO is something different).</p>
<p>Please advise, can I use the term 'entity' in English-written documents (e.g. 'list of newly built entities', 'entity # 1'), or it is better to use such terms as works, structure, facility, unit, site, plant? So that it be understandable for English-speakers in construction environment. (Note that the words used do not need to conform to ISO 12006-2; I have just used it as a reference.)</p>
<p>What terms have you really met and used for such cases?</p>
|
|civil-engineering|construction-management|
|
<p>I am a native English speaker and practising civil engineer who had never heard of ISO 12006-2 until this question, and if you had talked about "Construction Entities" I wouldn't have had a clue what you were on about.</p>
<p>Consider using "4 pump stations and 3 water mains", as this is clear. </p>
<p>If you want a word which covers all 7 "construction entities" I could comprehend:</p>
<ul>
<li>7 separate <strong>works</strong></li>
<li>7 separate <strong>components</strong></li>
<li>7 separate <strong>subsystems</strong> (<em>though this may be more suitable for relatively complicated mains)</em></li>
</ul>
<p>I don't think "structure" would be appropriate for a water main, and your other suggestions sound even less appropriate.</p>
|
12032
|
Construction Entity and Construction Complex
|
2016-10-26T22:24:51.253
|
<p>I was reading a paper about a "Controller Design for Temperature Control of Heat Exchanger" and I couldn't understand how to model the valve used in the system.</p>
<p>No need to read the paper as I will wrap up the whole issue:</p>
<p>Given:</p>
<ul>
<li>capacity of control valve is 1.6 kg/sec</li>
<li>time constant is 3 sec</li>
<li>valve input is pressure varying from 3 to 15 psi</li>
</ul>
<p>The resulting transfer function is:
$$G(s)=\dfrac{0.13}{3s+1}$$</p>
<p>It is obvious that they considered the valve as a first order transfer function.The gain $K_p$ was calculated by: $\dfrac{1.6\text{ kg/s}}{15-3\text{ psi}}=1.6/12=0.133$. Time constant = 3 seconds and that's it.</p>
<p>But the problem is: shouldn't a step input of 15 psi output a signal (1-exponential) with a final value of 1.6 kg/sec. But that transfer function won't actually output that. It still needs some kind of a shift. So am I missing something? Is the model wrong?</p>
<p>I also simulated the transfer function response with MATLAB SIMULINK along with a suggested alternative that probably solve the offset thing:</p>
<p><a href="https://i.stack.imgur.com/Abl1d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Abl1d.png" alt="SIMULINK MODEL"></a></p>
<p>The resulting waveform:</p>
<p><a href="https://i.stack.imgur.com/QxOOv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QxOOv.png" alt="Waveforms"></a></p>
<p>The yellow waveform is the input step function</p>
<p>The blue waveform is the output of the transfer function of the valve according to the paper.</p>
<p>The yellow waveform is the waveform with an offset.</p>
<p>Problem with the blue waveform: wrong final value.</p>
<p>Problem with the yellow waveform: starts from a negative value (-0.4 kg/sec)</p>
<p>So what is the correct model?</p>
<p>If necessary, <a href="http://www.wseas.org/multimedia/journals/control/2014/a125703-288.pdf" rel="nofollow noreferrer">here's the pdf file</a>.</p>
|
|modeling|
|
<p>The offset needs to be in the input and it's value needs to be 3.
<a href="https://i.stack.imgur.com/RWRRT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RWRRT.png" alt="enter image description here"></a></p>
<p>Then with the 15 psi input, the final value is 1.6 kg/sec.
<a href="https://i.stack.imgur.com/T9YUL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/T9YUL.png" alt="enter image description here"></a></p>
|
12040
|
Control System: offset issue with a simple valve model
|
2016-10-28T01:01:49.823
|
<p>What factors would degrade a spring's life span other than stress applied?</p>
<p>Will high cycles (e.g. 100k load-unload continuous cycles) and high speed (100 ms/cycle), affect the spring constant (k)? What does it means if the spring constant decreases or increases after 1 million cycles (does it means the spring has reached its lifespan, or degrade, or...)?</p>
<p>Consider the spring is under applied appropriate force.</p>
|
|mechanical-engineering|springs|
|
<p>As mentioned in the comments spring designs and materials can vary widely to it is difficult to generalise.</p>
<p>However if we consider steel springs which are very common then there are a few factors to consider. </p>
<p>Firstly if properly specified a spring should never exceed its elastic limit, however if this does happen then its rate may very well change due to work hardening and early failure is quite likely. </p>
<p>For a spring operating within specification then <a href="https://en.wikipedia.org/wiki/Fatigue_(material)" rel="nofollow">fatigue</a> is one possible cause of failure this occurs over a large number of cycles within the elastic limit as small local stress concentrations which cause local overloading and gradual crack growth untill the crack reaches a critical length and then propagates very rapidly. The gradual crack growth will have a small effect on the performance of a part as it will reduce its cross section slightly. </p>
<p>Specific materials will have a fatigue limit, this is stress at which a large number of cycles will lead to failure. Some materials have a fatigue limit close enough to the yield stress that fatigue does not need to be considered, in others the limit is so low that any practical component will be subject to fatigue eventually. </p>
<p>Fatigue failures are predicted statistically so you might say that a particular spring has a 99% chance of not failing before 100k cycles and a 50% chance of failing after 500k cycles based on experimental data and empirical formulae so the nominal design life before replacement is a compromise between the consequences of premature failure and the cost of replacement. </p>
<p>Another problem is creep this is plastic type deformation which occurs below the elastic limit when constant loads are applied for long periods. </p>
<p>Both creep and fatigue are generally calculated from empirical data specific to a specific material, manufacturing process and component class. </p>
<p>Another consideration is environmental corrosion and degradation. Metal springs can suffer from corrosion in certain conditions and polymers may break down over time due to UV light and reactions with environmental chemicals. </p>
|
12049
|
What affects the spring constant or spring life spans?
|
2016-10-28T08:26:27.260
|
<p>I have a structure with a lot of interconnecting column and beam frame elements. I want to compute the displacement, member forces/moments via Finite Element Method.</p>
<p>The problem now is that I don't know how to properly incorporate moment release at beam end. I understand that moment release means that at the beam's end, the bending moment is $0$, but I don't know how I can incorporate this piece of information ( as a boundary condition? Or as a form of penalty function, or...?) into my FEM model. </p>
<p>I know how to incorporate support into my FEM model though; I can just make the relevant appropriate stiffness terms very big. But not moment release. </p>
<p>Any idea how moment release can be formulated in finite element terms to achieve the desired end results?</p>
<p><strong>Note:</strong></p>
<p>This question is different from the one that I asked <a href="https://engineering.stackexchange.com/q/10864/3353">here</a>, because that one asked about mathematical formulation ( more on conceptual level), whereas this one asks about explicit FEM formulation. These two maybe related, but to go from previous question to this one is by no means trivial, and the connection not easy to see. </p>
|
|structural-engineering|structural-analysis|finite-element-method|
|
<p>Gere & Weaver address the need to place a hinge (end-release of rotational D.O.F.) at either end of a continuous beam element on pages 423 and 424 of 'Analysis of Framed Structures' and they have Tables 6-1 and 6-2 to illustrate their method. They imply that the member stiffness matrix modified in this manner can be assembled directly into the global (structural) stiffness matrix. Of course, the extra degree of freedom is not accounted for in this solution thus you must back-solve from the opposite end of the beam to present the discontinuous slope on the other side of the hinge to the user.</p>
<p>And there is a glaring problem with their solution in that zero elements on the pivot of the member stiffness matrix cannot be handled in traditional stiffness inversion techniques, generating instability messages of "zero pivot element". My personal solution was to insert 0.000001 value of stiffness on the pivot element where 0.0 would have been assigned when forming the modified member stiffness matrix.</p>
|
12050
|
How to incorporate moment release at beam end in FEM formulation?
|
2016-10-28T13:47:20.667
|
<p>We are looking to connect a non spinning coolant pipe to a spinning drill pipe. The connector needs to spin on the end with the drill pipe but be stationary on the end with the coolant pipe. Is there some type of bearing you know of or some other connector?</p>
|
|statics|dynamics|bearings|
|
<p>I think OP is asking about transferring fluid (coolant) through a rotating joint. </p>
<p>These kinds of joints are called <a href="https://en.wikipedia.org/wiki/Rotary_union" rel="nofollow">rotary unions</a> or swivel joints. One end of the rotary union is stationary, the other end may rotate continuously, and fluid passes through the joint, which is sealed even when the joint is in motion. </p>
|
12053
|
What kind of connection allows rotation at only one end?
|
2016-10-28T15:02:58.190
|
<p>What would be some common applications involving high friction and low wear rates? for example, gear-driven mechanisms which require high friction between the gear teeth but ideally low wear between them.</p>
|
|mechanical-engineering|applied-mechanics|friction|mechanisms|tribology|
|
<p>As mentioned in the comments you don't want high friction between gear teeth, but if you're asking about the materials who exhibits higher friction with low wear rate, it depends on several parameters, such as, the nature of the two counterfaces, the environment, the applied load, the hardness, velocity, etc
In general there's cubic crystal oxides, with a high hardness above 10GPA whom with sliding against certain material, exhibits highr friction coefficient with a loaw wear rate, we take an example of AL2O3.
Best regards</p>
|
12055
|
Applications with combination of high friction and low wear?
|
2016-10-28T16:41:08.630
|
<p>I have 3 programable logic controllers and other one which isn't part of my solution but I can access usefull data via modbus. I've setup windows service on a remote server to gather data and store it in a SQL Server.
But the problem is if the Internet fails? So I'm searching for a hardware data logger solution.
I saw a schneider scadapack but I think it doesn't support modbus as client.
Everything I'm seeing is hardware with I/O for digital and analog input/outputs but I have already that in PLCs. What I need now is to gather data from the PLCs via modbus RTU/TCP/UDP. After the service detects connection errors it registers the begin date and end date. After it detects PLCs are reachable again it downloads the usefull data that it missed from the hardware data logger solution.</p>
<p>Which solutions support this?</p>
<ul>
<li>Yokogawa MW100</li>
<li>Schneider scadapack models</li>
</ul>
<p>Both this are data acquisition systems via I/O and probably not via modbus.
I'm lost. Help?</p>
|
|data|
|
<p>Ice got experience with horner apg and mt-1xx/2xx by inventia.pl. Personally I've did data collecting from twido (modbus) on mt-151 (inventia, sdcard). You can did same thing on any horner device.</p>
|
12056
|
Data logger via modbus hardware solution
|
2016-10-29T18:50:19.300
|
<p>Generally, in a pressure drop control valve that uses a piston and a spring (for example, in jet engines to control fuel flow), there is always potential for error in terms of the height of the spring. As we know, a shim is usually used to fix this. However, I was wondering if there are alternatives to such a procedure. Are there other ways to correct the height on a valve spring, or perhaps produce the spring in a way that reduces the need for a correction method like shimming?</p>
<p>I was thinking maybe laser correction could be a viable method for correcting a spring that is off by a few thousands of an inch? It is extremely accurate, but then the main disadvantage is cost. Are there other methods or technologies that exist or are perhaps being worked on at the moment?</p>
|
|pressure|measurements|aerospace-engineering|flow-control|
|
<p>Shimming is always attractive for fit adjustment as it is fairly straightforward to do ie you can measure any gap with a feeler gauge and select the appropriate thickness of shim to suit. </p>
<p>The downside of manufacturing parts to fit exactly is that you then need to match individual parts to their location throughout their life and you then need a different system if you want to compensate for wear. Equally if you need a replacement part you need to report the exact size you need and get it made to fit as opposed to just ordering a standard part and having an existing known method to compensate for any variation. </p>
<p>Again if you know that the springs will always be a bit short you know you will always be filling some gap and its just the gauge of the shim which varies however if you are aiming for an acceptable fit out of the box there is always a risk that it will be slightly too long in which case a shim will no longer fix the problem.</p>
|
12069
|
Alternatives to shimming valve springs
|
2016-10-29T19:07:29.040
|
<p>Assume I have a system comprising of an airfoil <strong>A</strong> immersed in a fluid.</p>
<p>Further assume, that the airfoil is affixed via two bolts at one end, and is untethered at the other end. </p>
<p>The system can be parametised as follows:</p>
<ul>
<li><em>S</em> is the span of the airfoil</li>
<li><em>C</em> is the chord of the airfoil</li>
<li><em>V</em> (vega) is the relative velocity of the fluid passing over the airfoil</li>
<li>$\rho$ is the density of the fluid in which the airfoil is is immersed</li>
<li>$\theta$ is the current angle of attack of the airfoil (in radians)</li>
<li><em>L</em> the distance between two bolts affixing the airfoil at one end (< <em>C</em>)</li>
</ul>
<p><strong>[[<em>Question</em>]]</strong></p>
<p><strong><em>Given a new angle $\theta_i$, what would be the formula for calculating the force required to be exerted on the two bolts to change the current angle of attack of the airfoil from $\theta$ to $\theta_i$? (assuming all other variables held constant)</em></strong></p>
<p><strong>[[Notes]]</strong></p>
<p>From the lift characteristics of an airfoil, I think its fair to assume that a greater force will be required to be exerted as the angle of attack increases (up until the stall angle)</p>
<p>Additionally, since the force required is likely to be monotonically increasing (up until stall angle), I would prefer if the function actually returned the supremum of the the forces required to increase the angle of attack from:</p>
<p>$\theta$ $\rightarrow$ $\theta$ + $\delta$$\omega$</p>
<p>where $\delta$$\omega$ is the change in radians divided into an infinitesimal number of steps.</p>
<p>Ideally, the formula should derived from first principles (or the answer be provided as pseudocode for an algorithm), so that I can follow the logic, and apply it to an airfoil of non-rectangular shape.</p>
|
|mechanical-engineering|fluid-mechanics|applied-mechanics|aerodynamics|
|
<p>As per "NASA Aeronautics And Space Administration" the thin foil lift equation is $$L = Cl * A * .5 * r * V^2 $$ </p>
<p>Cl is the lift coefficient and in small angle range is directly related to angle of attack, multiplied by other factors. They have a java app here <a href="https://www.grc.nasa.gov/www/k-12/airplane/foil3.html" rel="nofollow noreferrer">FoilSim app</a> Which is similar to what you seem to be asking. You have to set the security of your computer Java to let this app run. They offer help to set it up. </p>
<p>As far as your model of the wing and its attachment to support via two bolts, it is not practical and minimum number of bolts required would be three non-collinear bolts to turn yor hinge connection into a cantilever connection. </p>
|
12071
|
Formula (or heuristic) for calculating force required for incremental change in angle of attack of an airfoil
|
2016-10-29T22:38:26.630
|
<p>I am having issues building a recreation of Focaults Pendulum (basic, inexpensive, undergrad project). At this point, I am going to purchase aircraft cable or suspension cable unless someone can recommend a slightly flexible rod. </p>
<p>I am going to use a 32oz plumb bob because I can't find a symmetrical spherical bob unless someone has suggestions. Last, if anyone has any recomendations on a frictionless or low friction pivot (360 degrees and pendulus) that would be amazing.</p>
|
|mechanical-engineering|structural-engineering|friction|experimental-physics|building-physics|
|
<p>There are two simple options for the bearing </p>
<p>The first is to suspend the bob from a flexible cord passing through a fairly tight grommet so it will maintain a constant point of contact in all directions. </p>
<p>Another option is to get a large ball bearing, ideally plastic or unhardened (or at least not through hardened) steel and then drill and tap it for an attachment point for your pendulum (or weld a nut to it in which case 316 stainless is a good option. </p>
<p>You can then mount this in a plate with a hole of something like 95% of the diameter of the bearing, for this slip-on pipe flanges are convenient. To get the best possible bearing surface fit you can lap the ball into the join with valve grinding paste or similar. </p>
<p>The example below is a slightly more complex version and can be locked made from a 50mm diameter 316 stainless ball bearing and a 1/1/2" bsp pipe flange. These flanges are failry thick but if you chamfer the top edge of the hole it gives you a bit more clearance and it can act as a reservoir for grease. </p>
<p><a href="https://i.stack.imgur.com/nt9ek.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nt9ek.jpg" alt="enter image description here"></a> </p>
<p>Edit to address questions in the commenst : </p>
<p>1) For a grommet or collet type support monofilament rather than braided wire might be best as the tension in braided wire might mean it behaves slightly differently in differnt directions. </p>
<p>2) For the ball bearing option it might be easiest to weld a nut to the bearing (or drill and tap for a stud) and thread the end of a round bar to fit. Idally you want the connection to the ball to be fairly rigid or there is little point using the ball at all. </p>
<p>3) There is probably no advantage in going for a very heavy ball as more weight will only increase the friction at the bearing surface having said that a larger ball give syou a larger range of movement. In my example a 50mm ball in a 48mm beveled hole gives around 45 degrees of motion in all directions from vertical. With this approach you are not relying on the felxibility of a 'string' of provide the range of mevement so you have more options in how you attach the bob to the bearing whihc in turn should allow you to use a heavier bob while keeping bearing friction and damping to a minimum. </p>
<p>4) I would guess that the magnetic donut is providing an impulse to the pendulum to compensate for friction losses and keep it swinging indefintely. This requires a coil and a modified ocsillator circuit...but that probably merits a new question. </p>
|
12073
|
Recreating Focaults Pendulum
|
2016-10-30T21:14:05.357
|
<p>Please help, what is the correct term for:</p>
<ol>
<li><p>The bottom/foot part of building or structure which carries loads from the building/structure and transfers the loads to the soil/earth/ground beneath it?</p></li>
<li><p>The soil massive to which said load from the structure is transferred.</p></li>
</ol>
<p>In my native tongue, the two meanings have different terms, however, in dictionaries they are named both base and/or foundation. The question is to distinguish the part of structure from the soil.</p>
|
|structural-engineering|civil-engineering|structures|geotechnical-engineering|terminology|
|
<p>As was said in the other answers, the part in contact with the soil is the "foundation", or perhaps "footing" (a specific type of foundation, but there are also piles, drilled shafts, rafts - <em>all</em> of these are foundations). </p>
<p>The soil under a footing will usually be called either the "bearing surface", or "bearing strata". However, this would not apply in the case of a deep foundation such as a pile foundation unless you are specifically talking about end-bearing piles. </p>
<p>Other terminology you may hear and find useful in different situations: </p>
<ul>
<li>soil matrix, or soil profile</li>
</ul>
<p>This refers to the layers of soils at some location, such as you will see in a boring log. </p>
<ul>
<li>bearing capacity</li>
</ul>
<p>This is a wildly variable term that very often means different things depending on who says it. When a structural engineer asks "What is the bearing capacity?", they are almost always asking for an ALLOWABLE bearing capacity based on some limitation for settlement (often "1 inch or less"). However, bearing capacity also means the amount of stress a soil can experience before undergoing some kind of shear failure. These are related, but totally different things. Always be sure you understand what is being talked about when people start throwing around "bearing capacity". </p>
|
12085
|
Base or foundation?
|
2016-10-31T00:42:38.170
|
<p><a href="https://i.stack.imgur.com/5g9sS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5g9sS.png" alt="enter image description here"></a></p>
<p>Everytime I try this question I keep getting 93.75 mm. I was just wondering where I'm going wrong. Here's my work: </p>
<p>$$
BC=\dfrac{FL}{AE}=\dfrac{1500000\text{ N}\cdot 1250\text{ mm}}{100000\text{ mm}^2*200\text{ MPa}}=93.75\text{ mm}
$$
For deflection to be in $\text{mm}$, all units have to be in $\text{N}$ and $\text{mm}$, right?</p>
|
|mechanical-engineering|
|
<p>You are correct. The book probably meant $E = 200\text{ GPa}$, which is frequently adopted for steel.</p>
<p>To confirm, I made a simple little model:</p>
<p><a href="https://i.stack.imgur.com/wDrYN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wDrYN.png" alt="enter image description here"></a></p>
<p>You're probably aware of this, but just to make it explicit: your math only worked because $200\text{ MPa} = 200\text{ N/mm}^2$.</p>
|
12087
|
Deflection problem confusing units?
|
2016-10-31T08:55:30.163
|
<p>I was just wondering for this question if you can see anything wrong with it. The answer in the textbook is 220MPa but i get 193MPa</p>
<p><a href="https://i.stack.imgur.com/bw6tV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bw6tV.png" alt="enter image description here"></a><a href="https://i.stack.imgur.com/Bffaa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bffaa.jpg" alt="enter image description here"></a></p>
|
|mechanical-engineering|
|
<p>The correct answer is 193 MPa as you have written. I have used <a href="https://skyciv.com/beam-calculator-features" rel="nofollow noreferrer">SkyCiv Beam</a> software to confirm this. It's highly unlikely you are both wrong. The shaft is essentially a beam so it's fine to use beam software for this.</p>
<p><a href="https://i.stack.imgur.com/HPmxh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HPmxh.png" alt="SkyCiv Beam used to determine bending stress of shaft"></a></p>
<p><a href="https://i.stack.imgur.com/21GKh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/21GKh.png" alt="Bending stress profile in SkyCiv"></a></p>
|
12089
|
Maximum Bending Stress
|
2016-10-31T10:11:31.097
|
<p>I am thinking of building a prototype for a solar generator using a <a href="https://en.wikipedia.org/wiki/Fresnel_lens" rel="nofollow">fresnel</a> lens to concentrate sun rays to a relatively small area, for the purpose of harnessing the thermal energy. </p>
<p>The problem with this is that the temperature at the concentrated point, <a href="https://www.quora.com/Thermodynamics-What-is-the-maximum-temperature-a-1-meter-square-fresnel-lens-can-reach-at-its-focal-point" rel="nofollow">temperatures as high as 3000F</a> are possible.</p>
<p>What (preferably inexpensive) material can be placed at the focal point of the lens, to transfer the heat away to do some useful work?</p>
|
|materials|thermodynamics|heat-transfer|solar-energy|
|
<p>Crucial to this is how you intend to exploit this energy as this will be very significant in how quickly and uniformly heat is extracted from the target. </p>
<p>3000F is outside the working range of most readily available engineering materials and if you want something which can withstand that sort of temperature outright you are in the realms of things like engineering ceramics, carbon composites and tungsten. </p>
<p>Having said that as long as you are removing heat quickly enough you should be able to keep the temperature in a sensible range indeed many heat engines have components which are exposed to flame temperatures in excess of their melting points. A gasoline engine will break pretty quickly if the cooling system fails. </p>
<p>So a logical approach is to use a heat engine with a working fluid with good heat transfer properties here steam power is an obvious choice and <a href="https://youtu.be/uJGpbvvJA2I" rel="nofollow">solar steam engines</a> do indeed exist (note that this is an illustration rather than a reccomendation of how to do it in detail).</p>
<p>Obviously any sort of boiler or steam power is fraught with hazards and not something which should be attempted without taking detailed and specialist engineering advice. There may also be leagal requirements for inspection and certification of boilers whcih apply. </p>
<p>As mentioned in the video answer simple balck steel pipe should be adequate as long as you have enought coolant flow and has the advanatge that there are existing standards for steam pipe and the associated fittings and it is raltiavely inexpensive and straightforward to work with. </p>
<p>This approach has the advantage that you are using the working fluid to directly take heat from the target of the lense with minimal intermediate steps between the heat source and the engine. </p>
<p>Also steam power is a mature and well developed technology so if you are more interested in eveluating the potential output of the lense you could just run it as an open system and measure the energy trasnfer to the water, avoiding the hazrds of a pressurised steam system. </p>
|
12091
|
What material (preferably relatively inexpensive), can I use for absorber in a fresnel solar concentrator
|
2016-10-31T18:38:36.077
|
<p>I am doing research on objects frozen with liquid nitrogen then seeing how they thaw.</p>
<p>I need to monitor objects temperatures as they thaw. Ideally getting the variance of the external and internal temperatures.</p>
<p>What strategy or equipment would you recommend to get this data.</p>
<p>Thank you.</p>
|
|temperature|
|
<p>You can get <a href="http://www.thermometricscorp.com/thermocouple.html" rel="nofollow">thermocouples</a> which are rated for cryogenic temperatures. These produce a voltage whcih can be calibrated to temperature and so are faily easy to connect to digiotal or analogue data logging equipment. </p>
<p>They are essentially just a wire so they are easy to embed in things and they can be fitted to various probes and pads for differnt mounting configurations. </p>
|
12098
|
Lab Equipment for Monitoring Sub Zero Temperatures
|
2016-11-01T04:41:52.447
|
<p>I am trying to do this question and I have everything but one. </p>
<p><a href="https://i.stack.imgur.com/XRQIE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XRQIE.png" alt="enter image description here"></a></p>
<p>Solving for $A_x$, $x$ force at $A$.</p>
<p>$$\begin{gather}
F_x = 6\cdot20 + A_x \\
A_x= -120\text{ kN}
\end{gather}$$</p>
<p>Is this correct?</p>
<hr>
<p>Sign normally doesn't matter, if you're consistent, but it's causing an error in my working out the moment.</p>
<p>Evaluating it, I get -1852.5 which is incorrect. The answer is 1132.5.</p>
<p>If I change the sign of $6\cdot20$ to $-6\cdot20$, I get the ‘correct’ answer, -1132.5.</p>
<p>$$\begin{gather}
F_x = 20\cdot6 + A_x = 0 \\
A_x = -120 \\
M_D = -30\cdot7\cdot3.5+2\cdot E_y = 0 \\
E_y = 367.5 \\
F_y = D_y + E_y - 30\cdot7 = 0 \\
D_y = -157.5 \\
F_y = -D_y - 60 + A_y = 0 \\
A_y = 60-157.5 = -97.5 \\
\sum M_A = M_A + 6\cdot20\cdot3-60\cdot4-D_y\cdot11=0
\end{gather}$$</p>
|
|beam|
|
<p>A moment is defined as the cross-product of distance and force vectors:</p>
<p>$$\begin{align}
M &= r\times F \\
&= \det\left(\left|\begin{matrix}
i & j & k \\
r_x & r_y & r_z \\
F_x & F_y & F_z \\
\end{matrix}\right|\right) \\
&= (r_yF_z-r_zF_y)\hat{i} -(r_xF_z-r_zF_x)\hat{j} + (r_xF_y-r_yF_x)\hat{k}
\end{align}$$</p>
<p>In a 2D frame, $r_z = F_z = 0$, which simplifies this to $M = (r_xF_y-r_yF_x)\hat{k}$. So, a positive vertical force at a positive horizontal distance (to the right of the studied node) results in a positive moment. Meanwhile, a positive horizontal force at a positive vertical distance (above the studied node) results in a negative moment.</p>
<p>A simple way of thinking about it is that force components are positive to the right or upwards and distances are positive if a positive force would generate a positive (counter-clockwise) rotation.</p>
<p>In your last equation, you made the moment due to the distributed load along $\overline{AB}$ positive, when it should be negative. After all, a positive horizontal force at a positive vertical distance (above $A$) results in a negative moment.</p>
<p>It should, however, be:</p>
<p>$$\begin{align}
\sum M_A =& M_A \\
& - (3)\cdot(6\cdot20) \\
& + (4)\cdot(-60) \\
& + (14.5)\cdot(-7\cdot30) \\
& + (13)\cdot (F_{E,y})= 0 \\
\therefore M_A =& -1132.5\text {kN}
\end{align}$$</p>
<p>I personally hate writing equations like this, with signs in the middle of the multiplications, but it makes all the variables explicit. So, in this calculation:</p>
<ul>
<li>all the distances are positive (everything is either above or to the right of $A$)</li>
<li>the horizontal distributed load is positive</li>
<li>the vertical loads are negative</li>
<li>$F_{E,y}$ is whatever it will be (positive, in our case)</li>
<li>When calculating the moment due to vertical forces, the sign is positive</li>
<li>When calculating the moment due to horizontal forces, the sign is negative</li>
</ul>
|
12101
|
Is a force to the right positive?
|
2016-11-01T11:33:02.207
|
<p>I'm trying to simulate a compound action potential for calibrating research instruments. The image below is a small overview of the design.
<a href="https://i.stack.imgur.com/GW1Dm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GW1Dm.png" alt="enter image description here"></a></p>
<p>I've already completed the data acquisition from a live animal, and processed the data in MATLAB to make a nice, noise-less signal, with 789 values in 12-bit format. I then cloned the repository to the Pi using Git. Below is the Python script I've written on the RPi. You can skip to def main in the script to see functionality.</p>
<pre><code>#!/usr/bin/python
import spidev
from time import sleep
import RPi.GPIO as GPIO
import csv
import sys
import math
DEBUG = False
spi_max_speed = 20 * 1000000
V_Ref = 5000
Resolution = 2**12
CE = 0
spi = spidev.SpiDev()
spi.open(0,CE)
spi.max_speed_hz = spi_max_speed
LDAQ = 22
GPIO.setmode(GPIO.BOARD)
GPIO.setup(LDAQ, GPIO.OUT)
GPIO.output(LDAQ,GPIO.LOW)
def setOutput(val):
lowByte = val & 0b11111111 #Make bytes using MCP4921 data sheet info
highByte = ((val >> 8) & 0xff) | 0b0 << 7 | 0b0 << 6 | 0b1 << 5 | 0b1 << 4
if DEBUG :
print("Highbyte = {0:8b}".format(highByte))
print("Lowbyte = {0:8b}".format(lowByte))
spi.xfer2([highByte, lowByte])
def main():
with open('signal12bit.csv') as signal:
signal_length = float(raw_input("Please input signal length in ms: "))
delay = float(raw_input("Please input delay after signal in ms: "))
amplitude = float(raw_input("Please input signal amplitude in mV: "))
print "Starting Simulant with signal length %.1f ms, delay %.1f ms and amplitude %.1f mV." % (signal_length, delay, amplitude)
if not DEBUG : print "Press ctrl+c to close."
sleep (1) #Wait a sec before starting
read = csv.reader(signal, delimiter=' ', quotechar='|')
try:
while(True):
signal.seek(0)
for row in read: #Loop csv file rows
if DEBUG : print ', '.join(row)
setOutput(int(row)/int((V_Ref/amplitude))) #Adjust amplitude, not super necessary to do in software
sleep (signal_length/(data_points*1000) #Divide by 1000 to make into ms, divide by length of data
sleep (delay/1000)
except (KeyboardInterrupt, Exception) as e:
print(e)
print "Closing SPI channel"
setOutput(0)
GPIO.cleanup()
spi.close()
if __name__ == '__main__':
main()
</code></pre>
<p>This script almost works as intended. Connecting the output pin of the MCP4921 to an oscilloscope shows that it reproduces the signal very well, and it outputs the subsequent delay correctly. </p>
<p>Unfortunately, the data points are seperated much further than I need them to be. The shortest time I can cram the signal into is about 79 ms. This is due to dividing by 789000 in the sleep function, which I know is too much to ask from Python and from the Pi, because reading the csv file takes time. However, if I try making an array manually, and putting those values out instead of reading the csv file, I can achieve a frequency over 6 kHz with no loss.</p>
<p><strong>My question is this</strong></p>
<p>How can I get this signal to appear at a frequency of 250 Hz, and decrease it reliably from the user's input? I've thought about manually writing the 789 values into an array in the script, and then changing the SPI speed to whatever value fits with 250 Hz. This would eliminate the slow csv reader function, but then you can't reduce the frequency from user input. In any case, eliminating the need for csv.read would help a lot. Thanks!</p>
|
|electrical-engineering|digital-communication|
|
<p>Figured it out earlier today, so I thought I'd post an answer here, in case someone comes upon a similar problem in the future.</p>
<p>The problem with the internal delay between data points cannot be solved with sleep(), for several reasons. What I ended up doing was the following</p>
<ul>
<li>Move <strong>all</strong> math and function calling out of the critical loop</li>
<li>Do a linear regression analysis on the time it takes to transfer the values with no delay</li>
<li>Increase the number of datapoints in the CSV file to "plenty" (9600) in MATLAB</li>
<li>Calculate the number of points needed to meet the user's wanted signal length</li>
<li>Take evenly seperated entries from the now bigger CSV file to fit that number of points as closely as possible.</li>
<li>Calculate these values and then calculate the SPI bytes explicitly</li>
<li>Save the two byte lists, and output them <strong>directly</strong> in the critical loop</li>
</ul>
<p>The new code, with a bit of input checking, is below</p>
<pre><code>#!/usr/bin/python
import spidev
from time import sleep
import RPi.GPIO as GPIO
import sys
import csv
import ast
spi_max_speed = 16 * 1000000 # 16 MHz
V_Ref = 5000 # 5V in mV
Resolution = 2**12 # 12 bits for the MCP 4921
CE = 0 # CE0 or CE1, select SPI device on bus
total_data_points = 9600 #CSV file length
spi = spidev.SpiDev()
spi.open(0,CE)
spi.max_speed_hz = spi_max_speed
LDAQ=22
GPIO.setmode(GPIO.BOARD)
GPIO.setup(LDAQ, GPIO.OUT)
GPIO.output(LDAQ,GPIO.LOW)
def main():
#User inputs and checking for digits
signalLengthU = raw_input("Input signal length in ms, minimum 4: ")
if signalLengthU.isdigit():
signalLength = signalLengthU
else:
signalLength = 4
delayU = raw_input("Input delay after signal in ms: ")
if delayU.isdigit():
delay = delayU
else:
delay = 0
amplitudeU = raw_input("Input signal amplitude in mV, between 1 and 5000: ")
if amplitudeU.isdigit():
amplitude = amplitudeU
else:
amplitude = 5000
#Calculate data points, delay, and amplitude
data_points = int((1000*float(signalLength)-24.6418)/12.3291)
signalDelay = float(delay)/1000
setAmplitude = V_Ref/float(amplitude)
#Load and save CSV file
datain = open('signal12bit.csv')
read = csv.reader(datain, delimiter=' ', quotechar='|')
signal = []
for row in read:
signal.append(ast.literal_eval(row[0]))
#Downsampling to achieve desired signal length
downsampling = int(round(total_data_points/data_points))
signalSpeed = signal[0::downsampling]
listlen = len(signalSpeed)
#Construction of SPI bytes, to avoid calling functions in critical loop
lowByte = []
highByte = []
for i in signalSpeed:
lowByte.append(int(i/setAmplitude) & 0b11111111)
highByte.append(((int(i/setAmplitude) >> 8) & 0xff) | 0b0 << 7 | 0b0 << 6 | 0b1 << 5 | 0b1 << 4)
print "Starting Simulant with signal length %s ms, delay %s ms and amplitude %s mV." % (signalLength, delay, amplitude)
print "Press ctrl+c to stop."
sleep (1)
try:
while(True): #Main loop
for i in range(listlen):
spi.xfer2([highByte[i],lowByte[i]]) #Critical loop, no delay!
sleep (signalDelay)
except (KeyboardInterrupt, Exception) as e:
print e
print "Closing SPI channel"
lowByte = 0 & 0b11111111
highByte = ((0 >> 8) & 0xff) | 0b0 << 7 | 0b0 << 6 | 0b1 << 5 | 0b1 << 4
spi.xfer2([highByte, lowByte])
GPIO.cleanup()
spi.close()
if __name__ == '__main__':
main()
</code></pre>
<p>The result is exactly what I wanted. Below is seen an example from the oscilloscope with a signal length of 5 ms; 200 Hz. Thanks for your help, guys!</p>
<p><a href="https://i.stack.imgur.com/ZQNEU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZQNEU.png" alt="Oscilloscope reading"></a></p>
|
12105
|
Python delays on Raspberry Pi
|
2016-11-01T15:15:46.360
|
<p>I am currently designing a prototype. I have asked to use low boiling point liquid to run a steam turbine. So i am thinking of using liquid ammonia.</p>
<p>What are the complications involved when using vapor ammonia in a steam turbine. Is it possible to run a small steam turbine using ammonia vapor?</p>
|
|turbines|
|
<p>Yes. However Ammonia is a potentially toxic substance if it leaks, however it's odor is a tell tale sign of a leak.
Ammonia is hydrogen based and high temperatures can diffuse it which can cause hydrogen fatigue damage to steels and alloys. So can the nitrogen.....</p>
|
12108
|
Is it possible to use Ammonia vapor in steam turbine
|
2016-11-02T02:32:21.293
|
<p>I'm trying to find a battery that I can use for a project. Let's be hypothetical for a moment.</p>
<p>I have 5-8 of <a href="https://www.adafruit.com/products/1142" rel="nofollow noreferrer">these Adafruit Servos</a>, connected to this <a href="https://www.adafruit.com/products/1411" rel="nofollow noreferrer">Adafruit Servo Shield</a>. Leaving the Arduino itself out of the equation, how can I calculate the amperage draw that these motors and this shield will pull? And additionally, how can I use that with the Ah of a battery to calculate the time that the circuit could be on and in motion with the battery in question?</p>
|
|electrical-engineering|motors|
|
<p>The <a href="http://www.electronicoscaldas.com/datasheet/MG996R_Tower-Pro.pdf" rel="nofollow noreferrer">datasheet for the device</a> should always list a maximum or "stall" current. For this device, it's 2.5A. </p>
<p>You can use that for a worst-case estimate of battery life by multiplying it by the amount of time you want to run. For example, if you want 12 minutes of run time, then the battery needs to have</p>
<p>$$
(2.5\mbox{A})*(12\mbox{min})/(60\mbox{min/hr}) = 0.5\mbox{Ahr}
$$</p>
<p><strong>Per servo</strong> that you have on your device. That is, if you have 5 servos, then you need a battery with a capacity of 2.5 Ah to run for 12 minutes. If you have 8, then you need a 4 Ah battery to run for 12 minutes. </p>
<p>Again, this is a worst-case scenario, if you try to run your servo at stall current for a long duration then you'll probably wind up burning them out. </p>
<p>How much capacity you actually need depends on the the <strong>duty cycle</strong> at which you're running the servos. If you're running an excavator then you might actually be running them close to stall current frequently. If you're trying to position a camera then you won't be.</p>
<p>My advice to you would be to go get a battery pack, any battery pack, and use that to power your system. Then, run the system as hard as you think it would normally be run. Compare the run time you achieved with the run time you desire. Then, adjust the capacity of the battery pack accordingly:</p>
<p>$$
\mbox{Required Capacity} = (\mbox{Test Capacity})*\frac{\mbox{Desired Run Time}}{\mbox{Actual Run Time}}
$$</p>
<p>So, if you had a 1.75 Ah battery and it ran for 5 minutes, but you want it to run for 20 minutes, then you need (1.75)*(20/5) = 7 Ah battery pack. </p>
|
12112
|
Amperage draw for Servo
|
2016-11-02T06:04:46.167
|
<p>When a body travels through any fluid(like aircrafts and cars), due to viscosity of the fluid a boundary layer is formed around the surface of the body which separates the velocity of the fluid inside the boundary layer(near the surface of object) to velocity of the fluid.The boundary layer consists of three zones laminar, transient and turbulent zones respectively.</p>
<p>Is it compulsory for all the three zones to exist in a boundary layer? If we consider a small smooth surface, is it possible that the surface ends at the laminar zone of the boundary layer and there is no turbulent zone in boundary layer over the surface? And if it is possible will there be a low pressure zone at the end of the surface or something?</p>
|
|fluid-mechanics|aerodynamics|
|
<blockquote>
<p>Is it compulsory for all the three zones to exist in a boundary layer?</p>
</blockquote>
<p>in practice, yes.</p>
<blockquote>
<p>If we consider a small smooth surface, is it possible that the surface
ends at the laminar zone of the boundary layer and there is no
turbulent zone in boundary layer over the surface?</p>
</blockquote>
<p>IF you have a sufficiently smooth surface, and sufficiently slow velocity, the NSE simplify (I use the term loosely) into a PDE that can be solved analytically. (stokes flow around a cylinder or sphere, for example)</p>
<p>The fact that they can be solved analytically indicate that the flow is not turbulent! (again, this is theoretical)</p>
<blockquote>
<p>And if it is possible will there be a low pressure zone at the end of
the surface or something?</p>
</blockquote>
<p>I'd have to crack open my fluid mechanics books before I can give a definite answer</p>
|
12114
|
What happens when the surface ends at the laminar zone of the boundary layer?
|
2016-11-02T07:21:41.393
|
<p>I'm doing a project which simulates a car accident, with a cube block which has sensors on it crashing into a wall.
This object is on a 3 m track mounted on a linear bearing to reduce friction.</p>
<p>So this object needs to be travelling at a speed and it impacts a solid wall (with about 40 G of deceleration in that impact). This impact is what I am recording.</p>
<p>My question is, how can I get this object to travel at that speed? (which is about 35 to 45 mph, if my math isn't wrong that is)</p>
|
|mechanical-engineering|
|
<p>You can't really calculate the G-force for two rigid objects, as the collision is nearly a delta function. If you want to simulate the force on, say, the driver inside a car, then you have to model the rate of deceleration as the front of the car collapses -- which is strongly dependent on the total vehicle design. As an historical note, Volvo back in the 60s and 70s ran ads showing their car being dropped nose-first off a roof, and retaining the basic frame shape after hitting the ground. That's great for rigidity, but then everyone realized this just transferred the peak impulse deceleration straight to the occupants. After that, cars' designs were modified to dissipate as much energy as possible by completely destroying the front end up to the firewall. This greatly reduces the peak G-force on the passenger compartment.</p>
<p>So, back to your block-to-wall experiment: unless your block is collapsible, you're going to be doing the 1970s Volvo thing. You may want to modify your experiment in light of this.</p>
|
12116
|
Getting an object to travel at 35 to 45 mph
|
2016-11-02T11:31:46.443
|
<p>Three common terms used in leadscrews are lead, # of starts, pitch. There definitions are usually as follows:</p>
<ul>
<li>Lead usually refers to the amount of linear travel when the leadscrew
turns one revolution. </li>
<li># of Starts refers to the number of independent threads along the length of the leadscrew. </li>
<li>Pitch refers to the distance between threads. In leadscrew with one start, pitch = lead. With multiple starts pitch = lead / # starts. </li>
</ul>
<p>In many leadscrew catalogues, the equation listed for translating a load's linear mass into angular inertia is:</p>
<p>$ J_{load} = m\times({\frac{1}{2\pi p}})^2$</p>
<p>where:</p>
<p>$m$ is the mass of the load.</p>
<p>$p$ is the pitch in <strong>threads per inch</strong>.</p>
<p>My main issue here is that the units are <strong>threads per unit length</strong> instead of <strong>length per thread</strong>. The mix in nomenclature here is very confusing. In a leadscrew with one start, I can easily do $\frac{1}{lead}$ and plug that as $p$ into the equation. But what about when there are multiple starts? Do I still use $\frac{1}{lead}$, or $\frac{1}{\#ofstarts\times lead}$? What is the derivation of this equation?</p>
|
|mechanical-engineering|threads|
|
<p>Firstly, in reference to nomenclature: in <em>ISO</em> (metric) standards the pitch of a leadscrew is given in units of distance/revolution. However, <em>Unified</em> (inch series) standards, pitch is given in <em>TPI</em> (threads per inch). As you mentioned in your question, lead is related to pitch by <span class="math-container">$p = L/N$</span> where <span class="math-container">$p$</span> is the pitch, <span class="math-container">$L$</span> is the lead and <span class="math-container">$N$</span> is the number of starts.</p>
<p>Source: Juvinall & Marshek, <em>Fundamentals of Machine Component Design, 5th Ed.</em>, pp. 412.</p>
<p>Here's a derivation of the equation.
The equation relating linear load to torque for a lead screw is:
<span class="math-container">$$ T = \frac{W d_m}{2}\frac{f\pi d_m + L \cos{\alpha_n}}{\pi d_m \cos{\alpha_n} - f L} $$</span>
Where <span class="math-container">$T$</span> is the torque, <span class="math-container">$W$</span> is the linear load, <span class="math-container">$d_m$</span> is the mean diameter of thread contact, <span class="math-container">$f$</span> is the coefficient of friction, <span class="math-container">$L$</span> is the metric lead in distance/thread, and <span class="math-container">$\alpha_n$</span> is the thread angle measured in the normal plane.</p>
<p>Source: Juvinall & Marshek, <em>Fundamentals of Machine Component Design, 5th Ed.</em>, pp. 418.</p>
<p>Let us examine the simplified case of a square thread (<span class="math-container">$\alpha_n = 0$</span>). The above equation then simplifies to:
<span class="math-container">$$T = \frac{W d_m}{2}\frac{f\pi d_m + L}{\pi d_m - f L} $$</span></p>
<p>Values of <span class="math-container">$f$</span> tend to be around 0.08 to 0.20 for steel against cast iron and bronze, and can be even lower for ballscrews (same source as above). Therefore let us simplify further by assuming <span class="math-container">$f = 0$</span>. Then we get the following:
<span class="math-container">$$ T = \frac{WL}{2\pi} $$</span></p>
<p>Now in this case our load <span class="math-container">$W= ma$</span> (mass times acceleration). Acceleration is related to angular acceleration by <span class="math-container">$a = \frac{L}{2\pi}\frac{d^2\theta}{dt^2}$</span>. This gives:
<span class="math-container">$$ T = \frac{mL^2}{(2\pi)^2}\frac{d^2\theta}{dt^2} $$</span>
From the definition of rotational inertia (<span class="math-container">$T = J\frac{d^2\theta}{dt^2}$</span>) we arrive at the final answer:
<span class="math-container">$$ J = \left(\frac{L}{2\pi}\right)^2m$$</span>
In other words, the torque you need to apply to move a linear load is independent of the number of starts of the leadscrew. For a single start, and using the <em>Unified</em> standard definition of pitch we get:
<span class="math-container">$$ J = \frac{1}{(2\pi p)^2}m $$</span></p>
<p>The results of my derivation are supported by the catalogue I found that lists the calculation for the force-torque conversion for a lead screw (pp. 2, eq. 3 and the sample calculations on pp. 9) <a href="https://www.sdp-si.com/D810/PDFS/Ball%20And%20Acme%20Lead%20Screw%20Technical%20Info.pdf" rel="nofollow noreferrer">here</a>.
Note that they include the efficiency value <span class="math-container">$e$</span> which is used to approximate the friction in the system while simplifying the calculations. You should double check the equation given by whatever catalogue you are using.</p>
<p>Hope that helps.</p>
|
12118
|
Calculating Load Inertia in a Leadscrew System
|
2016-11-03T03:45:22.140
|
<p>On the image below is a <a href="http://www.kopemash.ru/products/2/76.html" rel="nofollow noreferrer">mining machine produced</a> by the Kopeysk Machine-Building Plant. I'm not sure how to call it in English. The Russian name is "prokhodchesko-ochistnoy kombain". </p>
<p>Multitran offers the translation "<a href="http://www.multitran.ru/c/m.exe?l1=2&l2=1&s=%EF%F0%EE%F5%EE%E4%F7%E5%F1%EA%EE-%EE%F7%E8%F1%F2%ED%EE%E9%20%EA%EE%EC%E1%E0%E9%ED" rel="nofollow noreferrer">heading and winning machine</a>", but I'm not sure: when I enter this term in Google, it finds predominantly Russian-based websites. </p>
<p>The machine is used for underground mining. This particular model is mentioned in a text I'm translating about an overpricing row between the Russian potash producer Uralkali and the machine-building plant. </p>
<p><a href="https://i.stack.imgur.com/wcnH8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wcnH8.jpg" alt="enter image description here"></a></p>
<p>I found <a href="http://www.kopemash.ru/en/products/68/367.html" rel="nofollow noreferrer">the company's description</a> of the machine:</p>
<blockquote>
<p>The «Ural-20R-00» heading-and-winning machine is designed for longwall operation in сhambers and driving of oval-arch shaped workings in potash ore seams with thickness of 3.0-3.7 m at dip angles of ± 12° and rock resistance to cutting of up to Аp=450 H/mm. </p>
</blockquote>
|
|terminology|mining-engineering|
|
<p>The machine is nothing like a longwall miner. Longwall mining machines are limited in how they operate. A wall of shields protects the face being cut while a shearing head runs along the face cutting coal, dropping coal onto a short conveyor belt running parallel to the face. That conveyor belt then drops the coal onto another conveyor belt running perpendicular to the face being cut. As the longwall face is cut, the shields slowly advance forward. Also, longwall miners cannot produce curved tops to excavation; they can only produce flat topped excavations.</p>
<p>The machine in the question is more like a <a href="http://alpinecutter.com/en/products/roadheaders.html" rel="nofollow noreferrer">road header</a>. They are used to <a href="https://www.linkedin.com/pulse/sany-developed-ebs630-roadheader-potash-mining-%E6%89%AC-%E6%9D%8E" rel="nofollow noreferrer">mine postash</a> and other soft stratified deposits like salt. If needed, they can produce excavations with curved profiles. The machine in question may have purposely been designed & manufactured to mine a particular type of potash.</p>
|
12129
|
What would you call this mining vehicle? (Ural-20R, used in mining)
|
2016-11-03T12:44:56.823
|
<p>The gas $(A)$ is diffusing into the solid sphere $(B)$, The sphere phase is assumed to be stagnant so we have no bulk movement, $(A)$ is being consumed in the sphere by a rate of $KC_A$ which $K$ is a kinetic constant and $C_A$ is concentration of $A$ in the sphere, $C_A$ in sphere as a function of $r$ (radius) is desired.</p>
<p>By the species continuity Eqn. we have :</p>
<p>$\frac{1}{r^2} \frac{d}{dr}(r^2N_A)=-KC_A$</p>
<p>By definition of $N_A$ we have:</p>
<p>$N_A=y_AN_t-D_{AB}\frac{dC_A}{dr}$</p>
<p>which $N_t$ is zero by the stagnant assumption. By substituting $N_A$ in the first Equation and differentiating we obtain:</p>
<p>$C_A''+\frac{2}{r}C_A'-\frac{K}{D_{AB}}C_A$=0</p>
<p>I just don't know how to solve this differential equation, maybe there's a need for a simplifier assumption that I didn't make, anyway I welcome any advice from you guys helping me find the answer.</p>
|
|chemical-engineering|
|
<p>Your ODE is: $$c''+\frac 2r c'-\frac KDc=0.$$</p>
<p>Assuming $r\neq 0$, use the substitution $c = \frac {f(r)}{r}$. After substitution you will obtain $f''-\frac KD f=0$. This ODE can be solved by hyperbolic sine and hyperbolic cosine functions or simple exponential functions.</p>
|
12134
|
Molecular diffusion into sphere
|
2016-11-04T01:42:04.383
|
<p>Given a 3.25" long, 0.125" diameter rubber rod with tensile strength 1200 psi supported on both ends, how much weight can you hung from the middle of it without breaking it? (This is not homework I am trying to design a PC case and I have no idea what I am doing.) How does this change if we use a 0.25" diameter rod instead?</p>
|
|mechanical-engineering|structural-engineering|structures|beam|statics|
|
<p>I searched and could not find any rubber material which has the same modulus of elasticity for compression and tension. Also rubbers swell under stress and creep( plastic behavior over time). </p>
<p>So the question does not have an analytical answer. However as a hypothetical case and just for illustration one can try to stablish a range of I, (second moment of area) by testing a short sample of the material under different compression stresses and set a lose curve fitting graph for the I by stablishing a new neutral axis and integrating y^2.da separately for top and bottom parts over and below the axis. And go from there.</p>
|
12143
|
Given the tensile strength of a rod, how do you calculate the max weight that can be hung from it?
|
2016-11-04T10:52:22.373
|
<p>There is this device I saw when I was a child at a big music store, that makes me curious.</p>
<p>It was meant to listen some CDs to try them out before the purchase.
You stand under a bell-shape kind of "speaker" to hear the music, but as soon as you step out you aren't able to hear anything anymore.</p>
<p>I've been searching the net, but probably I haven't used the appropriate keywords because I didn't find anything.
What's this device called, and how is does it work?</p>
<p>I'm interested in particular about the sound and how it's possible to transmit it only in a very limited area, and being able to hear it well only inside that little "circle"</p>
|
|power-electronics|sound-isolation|waves|
|
<p>Something like this?
<a href="http://www.browninnovations.com/sound-dome/single-localizer-sound-dome" rel="nofollow noreferrer">http://www.browninnovations.com/sound-dome/single-localizer-sound-dome</a></p>
<p>As others have said, it is a speaker with a parabolic reflector to direct the sound down at the listener. You can make your own fairly easily: <a href="http://www.instructables.com/id/umbrelAudio-Sound-Dome-Umbrella/" rel="nofollow noreferrer">http://www.instructables.com/id/umbrelAudio-Sound-Dome-Umbrella/</a></p>
|
12147
|
What device is this, and how does it work?
|
2016-11-05T02:46:39.877
|
<p>So I have the following circuit:</p>
<p><a href="https://i.stack.imgur.com/8VzpZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8VzpZ.png" alt="enter image description here"></a></p>
<p>And the first question is, if DV1 is pushed all the way to the right, what speed will the clamp cylinder attain and as it extends what will be the pressure at P1.</p>
<p>Similarly Once the clamp cylinder has become fully extended, if the button on DV2 is pressed, what will be the speed of the piston of the lift cylinder and the pressure at P1?</p>
<p>The specifications are as such:</p>
<ul>
<li>Clamp cylinder -> Bore = 20 mm diameter, rod of 15mm diameter and a stroke of 1m.</li>
<li>Lift Cylinder -> Bore = 200mm diameter, rod = 15mm diameter, stroke = 1m</li>
<li>Clamp cylinder has a resistive force of 2.5kN that opposes motion</li>
<li>Lift Cylinder moves a mass of 50kg and no resistive forces</li>
<li>The relief valve is set to 20 MPa</li>
<li>Pressure valve, PV2 is set to 10MPa</li>
<li>Pump -> 5mL/rev and 1440 RPM</li>
<li>no pressure losses in line or components or internal leakage</li>
</ul>
<hr>
<p>So my thoughts are;</p>
<p>The pump is pumping at 5 *1440 = 7200mL/minute.</p>
<p>I'm going to just ignore PV1 as it's just a pressure relief valve, so it will just limit the system to whatever it is set to (20MPa).</p>
<p>I think we can ignore PV2 and everything beyond for the first question too. Or since this flow goes back to tank, I think it may mean that the system can't be more than 10MPa?</p>
<p>Now I think we calculate the Area of the head of the cylinder which is A = (0.2/2)^2 * pi = 0.0314 m^2.</p>
<p>So from here I think we can just put it into the formula Vext = Qp / A.</p>
<p>So 7.2*10^-3 / 3.14*10^-2 = 0.2293 m/s</p>
<p>Is this correct, or do I have to subtract some flow that would go to the other parts of the system (towards flow 2 and PV2).</p>
<p>Then the pressure at head should be P = F/A. This is my main problem, as 2.5*10^3/0.0314 = 80MPa which is way higher than any of the possible answer choices. Not to mention higher than what PV1 is set for.</p>
<p>So where have I gone wrong and how can I solve these problems?</p>
|
|fluid-mechanics|
|
<p>So when DV1 is pushed to the right the flow will be acting on the head of the cylinder so we work out the head Area:</p>
<pre><code>A = PI * r^2
</code></pre>
<p>As you said however you used the radius of the lift clamp not the clamp cylinder so it should be 0.02 (20mm) divided by 2 squared times pi.</p>
<p>This will give A = 0.000314 m^2.</p>
<p>The next step is, as you said, using the formula:</p>
<pre><code>v = q/A
</code></pre>
<p>However your units are a bit messed up. If your <code>q</code> is in L/s (q = 5 * 10^-3 * 1440/60 = 0.12) then you need to convert it to m^3 by multiplying it by 10^-3 and then dividing it by the area:</p>
<pre><code>v = 0.12 L/s * 10^-3 / 0.000314 m^2
v = 0.382 m/s
</code></pre>
<p>Now the pressure will be:</p>
<pre><code>P = F/A
P = 2500/0.000314
P = 7.9 * 10^6 Pa =~ 8 MPa
</code></pre>
<p>Since this is less than the 10 for the sequence valve we won't have to go back and adjust the flow.</p>
<p>As to your second question, since the lift cylinder is fully extended the flow cannot go to the clamp cylinder anymore so all of it goes to the lift cylinder so the process is the same:</p>
<pre><code>v = q/A
v = 0.12 * 10^-3 / 0.0314
v = 0.00382 m/s =~ 4 mm/s
</code></pre>
<p>and same for pressure (with a slight difference)</p>
<pre><code>P = F/A
P = 50 * 9.8 / 0.0314
P = 15.6 kPa
</code></pre>
<p>Since P here is 15.6 kPa, which is less than the 10 MPa of the sequence valve, the pressure at P1 will be 10 MPa.</p>
|
12157
|
Calculating speed, forces and pressures when extending a cylinder
|
2016-11-05T14:28:07.443
|
<p>In this question my bending moment diagram (BMD) doesnt return to 0, what's wrong with it?</p>
<p>To get the bending stress $\sigma = My / I$ , so i have to get the M (maximum moment) first.</p>
<p>To get the maximum bending moment, I have to get it from the graph of bending moment. To get bending moment, i have to get the shear force from the shear force diagram (SFD).</p>
<p><img src="https://i.stack.imgur.com/81GVb.jpg" alt="">
<a href="https://i.stack.imgur.com/pCsqB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pCsqB.jpg" alt="enter image description here"></a></p>
|
|structural-engineering|
|
<p>Your shear diagram is correct.</p>
<p>Bending moment is equal to the integral of shear. Therefore, the point of maximum bending moment is the point of zero shear. I can see you calculated that point as 4.38, which is correct (in fact, it's 4.375, but rounding is fine). That is found by</p>
<p><span class="math-container">$$6\cdot\dfrac{175}{175+65} = 4.375\text{ m}$$</span></p>
<p>Now you know the point of maximum bending moment. You chose to calculate the value by integrating the shear diagram. That is equivalent to calculating the total area of the shear diagram up to that point, which is equal to</p>
<p><span class="math-container">$$175\cdot\dfrac{4.375}{2} = 382.8\text{ kNm}$$</span></p>
<p>Your mistake was that you calculated the area of the rectangle <span class="math-container">$(175\cdot4.375$</span>) instead of the triangle that describes the shear diagram.</p>
<p>Adopting the correct bending moment value, you can then correct the rest of it as well:
<span class="math-container">$$\begin{align}
M_B &= 382.8 - 65\cdot\dfrac{6-4.375}{2} = 330.0\text{ kNm} \\
M_C &= 330.0 - \dfrac{125+205}{2}\cdot2 = 0\text{ kNm}
\end{align}$$</span></p>
<p>And now, to check our work:</p>
<p><a href="https://i.stack.imgur.com/K0W8C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K0W8C.png" alt="enter image description here"></a></p>
<p><sub>Diagram obtained with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis program.</sub></p>
|
12165
|
Bending moment diagram from shear force diagram
|
2016-11-05T23:48:20.657
|
<p>In <a href="https://i.stack.imgur.com/AxBsb.png" rel="nofollow noreferrer">this example</a>, I suspect the selected A' (area) is wrong... From <a href="https://i.stack.imgur.com/bRnF0.png" rel="nofollow noreferrer">the notes</a>, A' is the "portion of the member's cross sectional area above the section where t is measured". But, in the example, we could see that the selected area is located to the right of the point where shear stress is calculated. Are the notes wrong? </p>
<p><strong><em>I believe the selected A' should look like <a href="https://i.stack.imgur.com/AfHQe.png" rel="nofollow noreferrer">this</a></em></strong> (refer to green shaded part). Am I right?</p>
|
|structural-engineering|
|
<p>The area A' is the excluded area for the cut under consideration. To get the shear stress at a location you "cut" the section there, and look at the area beyond the cut.</p>
<p>For the shear stress in the web at the top of the web, the whole of the flange is excluded (your third picture). For the shear stress in the flange adjacent to the web, only the part of the flange beyond that cut is excluded (your first picture).</p>
|
12175
|
shear stress at particular point in beam
|
2016-11-06T01:21:10.557
|
<p>For a isosceles triangle with base <span class="math-container">$b$</span> and height <span class="math-container">$h$</span>, the surface moment of inertia around the <span class="math-container">$z$</span>-axis is <span class="math-container">$\frac{bh^3}{36}$</span> (considering that our coordinate system has <span class="math-container">$z$</span> in the horizontal and <span class="math-container">$y$</span> in the vertical axis, and has its origin on the triangle's center of mass (which is at <span class="math-container">$\left\{\frac{b}{2},-\frac{h}{3}\right\} $</span> if you put your coordinate system's origin at the bottom left corner if the triangle).</p>
<p>I know that the formula for the moment of inertia around the <span class="math-container">$z$</span>-axis is <span class="math-container">$I_z = \int_A{y^2 dA}$</span>, but I cannot figure out how to derive the formula from that. How is it done?</p>
|
|structural-engineering|applied-mechanics|statics|
|
<p><a href="https://i.stack.imgur.com/Ybr3h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ybr3h.png" alt="enter image description here"></a></p>
<p>The infinitesimal area $\text{dA}$ is $2 z \text{dy}$.</p>
<p>The relationship between $z$ and $y$ can be got from the slope.
$$\frac{z}{y-\frac{2 h}{3}}=\frac{0-\frac{b}{2}}{\frac{2
h}{3}+\frac{h}{3}}$$</p>
<p>Solving, we get
$$z=\frac{b (2 h-3 y)}{6 h}$$</p>
<p>Thus
$$I_{zz}=\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 dA
=\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 2 z \, dy
=\int_{-\frac{h}{3}}^{ \frac{2 h}{3}} y^2 2 \frac{b (2 h-3 y)}{6 h} \, dy=\frac{b h^3}{36}$$</p>
|
12176
|
Derivation of the formula for the surface moment of inertia of an isosceles triangle
|
2016-11-07T16:50:53.833
|
<p>I believed that the flame propagates over the place where fuel contacts the oxygen. If it contacts a fule+O2 mixture, the mixture inflames as well, and so on and on and on until there is no more inflammable material available around. If there is a gas comes to the surface, the flame stops and starts burning at the gas-air contact zone. </p>
<p>But, <a href="https://diy.stackexchange.com/questions/102195/">as it was explained yesterday</a>, I can have the gas burning in my gas baking chamber because it gets mixed with air prior to getting into the chamber. This explain the mystery of having the flame in a sort of isolated chamber, but, what they refused to explain is that how it is possible that mixture does not blow up in the pipe as soon as you mix gas + air? Why don't you have the flame there, how do you prevent the mixture ignition until after it crosses the designed burner? How does the gas know how the burner/nozzle look like in order to inflame in the right place? My belief was that flame propagates in all directions over the fuel+oxygen mixture, so, it should propagate from the burner down the pipe until the gas-air mixer, right?</p>
<p>BTW, is it right that <a href="https://diy.stackexchange.com/questions/21504/why-is-there-an-opening-where-the-burner-connects-to-the-gas-line-on-a-gas-oven">a hole in a gas pipe sucks the air despite gas</a> in the pipe has pressure > 1 atm (it must be greater than 1 atm in order to be pushed out of the gas pipe instead of sucking the air into it)? How is it possible that gas flows from lower pressuire into the higher pressure area without violation of entropy principle?</p>
<hr>
<p>I have noticed that there is a similar question on Quora. It is even more infomred than mine assumption -- it asks why no flahback arrestors are installed in the kitchen ovens? There should be absolutely no chance for the flame to go upstream over the explosive mixture! Why? </p>
<p>I keep asking my question because highly appreciated <a href="https://www.quora.com/Why-dont-we-provide-flash-back-arrestors-for-cooking-gas-cylinders" rel="nofollow noreferrer">answer there</a> is unsatisfactory. It just says that we use propane cooking cylinders</p>
<p>in our kitchens. These cillinders have much higher pressure than the air oxygen, and, therefore, backpropagation is not possible, says the answer.</p>
<p>However, it prevents the <strong>backward propagation of GAS</strong> whereas safety issue is propagation of <strong>flame</strong>. how does it make the flame speed necessarily lower than the gas flow speed, what happens on the last days of cylinder exhaustion when pressure goes down and what about the centralized (natural) gas kitchens?</p>
|
|gas|safety|combustion|
|
<blockquote>
<p>I have noticed that there is a similar question on Quora. It is even
more infomred than mine assumption -- it asks why no flahback
arrestors are installed in the kitchen ovens? There should be
absolutely no chance for the flame to go upstream over the explosive
mixture! Why?</p>
<p>I keep asking my question because highly appreciated answer there is
unsatisfactory. It just says that we use propane cooking cylinders in
our kitchens. These cillinders have much higher pressure than the air
oxygen, and, therefore, backpropagation is not possible, says the
answer.</p>
<p>However, it prevents the backward propagation of GAS whereas safety
issue is propagation of flame. how does it make the flame speed
necessarily lower than the gas flow speed, what happens on the last
days of cylinder exhaustion when pressure goes down and what about the
centralized (natural) gas kitchens?</p>
</blockquote>
<p>I think this part of question is not addressed, so I'll give it a try.</p>
<p>Kitchen ovens are supplied with only one gas source (butane cylinder in my case). While pressure in cylinder is greater than atmospheric pressure there can be no flashback because cylinder and lines which supply it to the oven are filled only with butane. And they are filled only with butane as long as pressure in the cylinder is greater than atmospheric. Since there is no combustible mixture ignition is not possible there, nor is flame able to propagate. So there is no danger of flashback/explosion.</p>
<p>Once cylinder/line pressure drops close to atmospheric gas flow will be too weak to sustain flame on the exit line (partly explained in mart's answer), so you'll be forced to replace the cylinder far before air gets inside. And even if air somehow gets inside and mix properly (which is not likely because butane is heavier than air, so it will always remain at the bottom of the cylinder) I don't think there would be enough energy released to cause cylinder to explode.</p>
<p>Where I'm from there was always warning not to turn the cylinder upside down when depleted, because then remaining butane could escape and make room for air.</p>
|
12203
|
How do you prevent ignition of the gas-air mixture before it reaches the burner nozzle?
|
2016-11-08T09:09:54.240
|
<p>Since diesel runs lean, why is soot produced in diesel exhaust? </p>
<p>Soot is carbon particles due to incomplete combustion. But, since diesel runs lean with plenty of oxygen, why does this happen?</p>
|
|combustion|engines|diesel|
|
<p>One important reason is that diesel fuel had a high molecular weight compared to gasoline this means that is is more difficult to disperse as it forms liquid droplets as opposed to vapour and even more importantly there are many more intermediate reactions involved in complete combustion. </p>
<p>For example Hydrogen, H2 burns very easily in oxygen as combustion is essentially one reaction process, H2 and O2 are separated and recombine as H20, simple. </p>
<p>However if you have a complex hydrocarbon fuel with long chain branched molecules there are lots of intermediate steps to go through to fully break it down into Carbon dioxide and water. </p>
<p>In an IC engine this combustion process has a very limited time to take place as an ignition cycle may take a fraction of a second </p>
|
12212
|
Why is soot produced in diesel exhaust, since diesel runs lean?
|
2016-11-08T16:04:54.833
|
<p>The wikipedia article for the Darcy Weisbach pressure drop equation states that "In effect, the friction loss in the laminar regime is more accurately characterized as being proportional to flow velocity, rather than proportional to the square of that velocity: one could regard the Darcy–Weisbach equation as not truly applicable in the laminar flow regime."</p>
<p>What pressure drop equation would be more appropriate for flow with Reynolds numbers 1-100?</p>
|
|fluid-mechanics|
|
<p>Googling "pressure drop laminar flow" gives a Wikipedia link to the <a href="https://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation" rel="nofollow noreferrer">"Hagen-Poiseuille" equation</a> for laminar flow through circular pipes.</p>
<p>Does that answer your question?</p>
|
12219
|
Pressure drop in very low reynolds number flows
|
2016-11-09T22:49:58.057
|
<p>As we know, if the damping model only depends on the instantaneous velocity, it can be called viscous damping, for which the damping force $f = Cv$. However, if $C$ depends on the deformation as $C(u),$ is this still named viscous damping? And if the damping force is modeled as $f=C g(v),$ where $g$ is a function of $v$, do we call this nonlinear viscous damping? Thank you!</p>
|
|mechanical-engineering|structural-engineering|modeling|
|
<p>I think "viscous" refers to the physical effect that causes the damping - i.e. the viscosity of a fluid. The viscous forces might be a linear or nonlinear function of the velocity, or something more complicated which might include a function of the deformation. The terms "linear viscous damping" and "nonlinear viscous damping" are meaningful in that context.</p>
<p>A damping force like of the $f = Cg(v)$ could model something like frictional damping, where for example $C$ was a (negative) constant and $g(v) = v/|v|$, (i.e. for linear motion $g(v) = \pm 1$ depending on the direction of the motion) but calling it "nonlinear viscous damping" instead of "friction damping" seems rather pointless.</p>
|
12228
|
deformation-dependent damping
|
2016-11-10T00:38:22.347
|
<p>This is about navigable or irrigation water ways.</p>
<p>Examples:</p>
<ol>
<li><p>Navigable channel or navigable canal? (it is about approach water ways to sea ports, or artificial inland water ways).</p></li>
<li><p>Channel dredging or canal dredging? (it is about navigable water ways).</p></li>
<li><p>Irrigation channel or irrigation canal? (it is about water ways within open irrigation systems).</p></li>
<li><p>Drainage canal or drainage channel? (it is about drainage water ways cut in earth).</p></li>
</ol>
<p>Comment: in my native language (Russian), the terms canal and channel have the same translation in dictionaries.</p>
|
|civil-engineering|terminology|water-resources|
|
<p>I thought this would be easily answerable by looking at English dictionaries. But they're not as useful as I'd hoped.</p>
<p><strong><a href="http://www.dictionary.com/browse/channel?s=t" rel="nofollow noreferrer">Channel</a></strong>:</p>
<blockquote>
<ol>
<li><p>the bed of a stream, river, or other waterway.</p>
</li>
<li><p><em>Nautical.</em> a navigable route between two bodies of water.</p>
</li>
<li><p>the deeper part of a waterway.</p>
</li>
</ol>
</blockquote>
<p><strong><a href="http://www.dictionary.com/browse/canal?s=t" rel="nofollow noreferrer">Canal</a></strong>:</p>
<blockquote>
<ol>
<li><p>an artificial waterway for navigation, irrigation, etc.</p>
</li>
<li><p>a long narrow arm of the sea penetrating far inland.</p>
</li>
</ol>
</blockquote>
<p>Well, I disagree with both of those.</p>
<ul>
<li><p>A sea is not a "waterway", but a deeper, navigable route through it is definitely a channel; possibly this is covered by definition 2? I'm not sure.</p>
</li>
<li><p>A long narrow arm of the sea penetrating far inland is a river, not a canal. In my experience (and, to be honest, backed up by the "British dictionary" further down the webpage), a canal can <em>only</em> be a manmade / artificial waterway.</p>
</li>
</ul>
<p>So, to answer your questions:</p>
<ol>
<li><p>Navigable channel or navigable canal? (it is about approach water ways to sea ports, or artificial inland water ways). <em>If a navigable path through the sea it can only be a channel. If a navigable artificial inland waterway, it can be a navigable canal, but you can also call it a navigable channel. Navigable channel would therefore always be acceptable for this.</em></p>
</li>
<li><p>Channel dredging or canal dredging? (it is about navigable water ways). <em>As above</em></p>
</li>
<li><p>Irrigation channel or irrigation canal? (it is about water ways within open irrigation systems). <em>Either is acceptable, assuming it is man-made (seems likely for an irrigation system). Personally, I would lean towards "irrigation channel". According to <a href="https://books.google.com/ngrams/graph?content=irrigation%20channel%2C%20irrigation%20canal%20&case_insensitive=on&year_start=1980&year_end=2008&corpus=15&smoothing=3&share=&direct_url=t4%3B%2Cirrigation%20channel%3B%2Cc0%3B%2Cs0%3B%3Birrigation%20channel%3B%2Cc0%3B%3BIrrigation%20channel%3B%2Cc0%3B%3BIrrigation%20Channel%3B%2Cc0%3B.t4%3B%2Cirrigation%20canal%3B%2Cc0%3B%2Cs0%3B%3Birrigation%20canal%3B%2Cc0%3B%3BIrrigation%20Canal%3B%2Cc0%3B%3BIrrigation%20canal%3B%2Cc0" rel="nofollow noreferrer">google ngrams</a>, "irrigation canal" is more common, but not massively so.</em></p>
</li>
<li><p>Drainage canal or drainage channel? (it is about drainage water ways cut in earth). <em>This is similar to the irrigation question above. Again, I would lean towards "channel". For this one, <a href="https://books.google.com/ngrams/graph?content=drainage%20channel%2Cdrainage%20canal&case_insensitive=on&year_start=1980&year_end=2008&corpus=15&smoothing=3&share=&direct_url=t4%3B%2Cdrainage%20channel%3B%2Cc0%3B%2Cs0%3B%3Bdrainage%20channel%3B%2Cc0%3B%3BDrainage%20channel%3B%2Cc0%3B%3BDrainage%20Channel%3B%2Cc0%3B%3BDRAINAGE%20CHANNEL%3B%2Cc0%3B.t4%3B%2Cdrainage%20canal%3B%2Cc0%3B%2Cs0%3B%3Bdrainage%20canal%3B%2Cc0%3B%3BDrainage%20Canal%3B%2Cc0%3B%3BDrainage%20canal%3B%2Cc0" rel="nofollow noreferrer">google ngrams</a> agrees with me (just!).</em></p>
</li>
</ol>
|
12230
|
Channel or Canal?
|
2016-11-09T05:49:18.947
|
<p>in tthe notes , It's stated that the shear flow will be varies lineraly prependicular to the direction of shear force , V . But , in the phooto, it shoed that the shear flow will be varies parabolically prependicular to the direction of shear force ...
Which is correct ? the notes or the example in the photo ?<a href="https://i.stack.imgur.com/5LkSu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5LkSu.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/Who9B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Who9B.png" alt="enter image description here"></a></p>
|
|structural-engineering|
|
<p>You're getting something mixed up in your head here.</p>
<p>The shear force is travelling in the vertical direction (up/down).</p>
<p>The segment perpendicular to that shear force is the horizontal component (left/ right) AKA the top and bottoms of the I beam. In this section the shear flow is linear on the drawing.</p>
<p>The centre section is going vertically, therefore it is parallel to the shear; and it shows the shear flow acting parabolically.</p>
<p>To summarize; both the notes and photo are correct; something in your interpretation was just misunderstood.</p>
|
12231
|
shear flow changes
|
2016-11-10T12:28:42.490
|
<p>How can we find it with the current technologies? Can it done without touching the haystack? Any cool stuffs like X-ray or some penetrating technologies ?</p>
|
|electrical-engineering|
|
<p>I'd say X ray is the best bet. Or possibly ultrasound, metal detector (high powered).
Maybe since you never said the haystack has to remain intact, you could burn it down and search for the needle with a metal detector in the ashy remains. </p>
|
12235
|
Find needle in a haystack?
|
2016-11-10T17:09:39.270
|
<p>I need to accurately measure (+/-0.25mm) the bolt hole circle drilled into a heavy metal plate. There are three holes on one plate and nine holes on a second plate. The plates are approximately 300mm diameter. The holes are approximately 18mm diameter tapped M20 and the bolt hole circle is about 200mm diameter.</p>
<p>When known, I will create a mating part on the CNC mill and lathe. The finished part rotates at around 2000 RPM, so the accuracy of the machining and fitment will determine the vibration of the finished system.</p>
<p>How can I measure the hole circle? Tools? Technique? Calculations?</p>
|
|mechanical-engineering|machining|
|
<p>The bolt pattern should be a standard; if there are three bolt holes then there should be 120 degrees between each hole. If there are nine bolt holes then there should be (360/9) = 40 degrees between each hole. </p>
<p>Let's talk for a second about vibration. Vibration will exist if the centerline of your flange does not coincide with the centerline of the existing flange. While your bolt pattern <em>does</em> matter, it doesn't matter <em>that much</em>. <strong>If you are relying on your bolt pattern to perfectly align two pieces of rotation machinery you are doing it wrong.</strong> The machinist that attaches the two flanges together should have a means to shim one piece of equipment, or the other, or both.</p>
<p>You need to rely on the shims to bring the two centerlines together, both coincident and parallel. Again, if you're relying on the bolt pattern to do this for you, then that implies that the flange needs to be <em>perfectly</em> attached to the shaft, that the bearings have to be <em>perfectly</em> sized with no room for dimensional tolerance, the body has to be perfect, the mounting feet have to be perfect, and the sub-structure to which both pieces of equipment attach has to be perfect. <strong>You need to have room for adjustment where the machinery bolts down.</strong> You will <strong>never</strong> attain the degree of perfection required to count on a bolt pattern to mate two pieces of rotation equipment. </p>
<p>So, that said, where vibration <em>does</em> matter with regards to your bolt pattern is that your bolt pattern should be EVEN. If you have three bolts that are supposed to be 120 degrees apart, but two are 110 degrees apart and the others are 125 each, then you'll get vibration because the bolt pattern isn't rotationally symmetric. </p>
<p>So, my advice to you would be to get the bolt pattern dimensions from the existing flange, then <strong>ignore minor variations</strong> and make your new pattern as symmetric as you can achieve. If the existing flanges are threaded, then your holes are through holes. There needs to be some tolerance there for the bolt to be able to pass through, and the tolerance should also allow for the minor variations in hole placement in both parts. </p>
<p>The clamping force of the bolts holds the flanges together. The shims under the mounting feet of the machinery brings the centerlines together. The location of the holes matters for vibration only if they're not rotationally symmetric. The fit of the the flange doesn't have anything to do with vibration provided you have adequately clamped the two flanges together and you aligned the shaft centerlines to be coincident, and again, that's done by moving the equipment, not by adjusting the flange bolt holes. </p>
<p>I'm harping on this to try to impress on you that a flange with through holes is not, and should not be, a piece of high precision equipment. You might need the <em>face</em> to be very flat if you're using a gasket, and it should definitely be perpendicular to the shaft centerline, but the through holes are just through holes. If one flange is able to rotate relative to the other flange then you haven't torqued it well enough and/or you haven't used enough fasteners. If the flanges were able to have relative motion then you're <a href="https://en.wikipedia.org/wiki/Clutch_control#Riding_the_clutch" rel="nofollow noreferrer">"riding the clutch"</a> so to speak and can expect (very) premature joint failure. </p>
<p>So, all that said, here's how to calculate what the parameters <strong>should</strong> be for your circular bolt pattern. </p>
<ol>
<li>Find the angular distance between bolts, $\theta$, by dividing 360 degrees by the number of bolts. A 3-bolt pattern is 120 degrees between fasteners, 4-bolt is 90, 9-bolt is 40, etc. </li>
<li>For a threaded hole, thread a bolt into each of two adjacent holes. Not necessary if you can get the measurement tips of a pair of calipers into the holes, but it does make it easier. </li>
<li>Measure the largest outside-outside distance between the two bolts. Measure the smallest inside-inside distance between the two bolts. Average those numbers (add together and divide by two) and you get the center-center distance. </li>
<li>The bolting pattern's radius is given by:</li>
</ol>
<p>$$
r = \frac{\mbox{average distance}/2}{\sin{(\theta /2)}} \\
$$</p>
<p>Now you can put your blank flange on a lathe, find the center, measure out a radius $r$, mark that circle, and locate your through hole centers at the appropriate angular positions on that circle. If you're doing it by hand you could open a compass to the $\mbox{average distance}$ between bolt holes that you found, put the pointy end anywhere on the circle and sweep it, then move the pointy end to any sweep-circle intersection and sweep again. </p>
<p>Here's a graphic if that'll make it any clearer:</p>
<p><a href="https://i.stack.imgur.com/kHudn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kHudn.png" alt="Bolt pattern geometry"></a></p>
|
12240
|
Accurately measure a bolt hole circle?
|
2016-11-10T18:41:02.703
|
<p>If I take a battery that is "dead" and completely replace the electrolyte and maybe the terminals, would it be a "fresh" battery?</p>
<p>So instead of pulling up to a charging station in your electric car, plugging in, and waiting half an hour or more to be able to continue, could you pull up, suction the electrolyte out, pump some fresh stuff in, and be back up to a fully charged state? </p>
<p>Would the electrolyte have to be charged before pumping it in?</p>
|
|battery|
|
<p>One plate is lead, one plate is lead dioxide. Hand-waving a big: sulfuric acid dissociates in water, making a lot of hydrogen sulfate ions. The lead plate reacts with the hydrogen sulfate to make lead sulfate, one hydrogen, and two electrons. The lead oxide plate takes a hydrogen sulfate, three hydrogens, and one electron to (also) make lead sulfate. This means there are three lead reactions required to make sufficient hydrogens for the lead oxide reaction, but the lead oxide reaction only consumes one of the six electrons generated. That is, the process releases 5 electrons. </p>
<p>The process also generates a considerable amount of lead sulfate. This forms as a powdery coating over all the plates in the battery. Interesting tidbit: any mechanical shock (bumps in the road, etc.) can knock that powdery coating off. As the coating is no longer attached to a particular plate, it's not going to get "energized" when the charge reaction happens, and thus will not convert from lead sulfate back to lead or lead oxide. Eventually, enough of this powder accumulates on the bottom of the battery and will short two (or more) plates, resulting in a higher self-discharge rate. Eventually, the self-discharge rate gets high enough that the battery won't "hold a charge" and then it gets replaced.</p>
<p>So, anyways, if all you do is pump out the electrolyte, then you haven't done anything to address the fact that the lead plates in the battery have corroded away (to form lead sulfate). Replacing just the electrolyte doesn't charge a battery. You would have to replace the electrolyte AND the lead plates.</p>
<p>But, if you take a dead battery, and then replace the electrolyte and charge plates, then it's a new battery. The easier thing to do than cracking open the battery and replacing all the internals would be to just have an exchange program, like <a href="http://www.bluerhino.com/Tank-Exchange.aspx" rel="nofollow noreferrer">propane tank exchanges in the US</a>. Users are charged a fee which equates to the energy (gas, or in this example electricity) added to the "vessel" as well as a small service fee to clean/inspect/recondition the "vessel." </p>
<p>Personally, I think that's the way of the future for electric vehicles. It's just too hard to provide the electric power required to charge a battery at anywhere close to the rate gas vehicles refuel, which "charge" at a rate of megawatts - check energy density of gasoline and multiply it by 10 gallons per minute. </p>
|
12244
|
How much of a battery needs to be replaced for it to be a fresh battery?
|
2016-11-10T19:44:58.423
|
<p>I am trying to see how much I can reduce the torque on a rudder post by changing the foil shape. I think I need a rudder with a centre-of-effort further ahead than average but I don't know what rudder series to use.</p>
<p>Balancing the rudder (in a spade rudder style) would be much simpler (and more efficient) - however, there is a design requirement for a sharply angled skeg for this particular project. The rudder will be attached/trailing directly behind a skeg.</p>
|
|mechanical-engineering|fluid-mechanics|aerodynamics|naval-engineering|airfoils|
|
<p>As a Naval Architect, this is a question that I am often asked. The centre of effort is very close to the geometric centre of the spade as you know. </p>
<p>Is there a fixed skeg or keel immediately in front of the rudder? That affects the hydrodynamics significantly because the hydrofoil would then have an effective length that is the skeg width (aligned with the fluid flow) plus the rudder chord length. While not precise, this approximation will allow a better calculation of Rn (Reynolds Number) to calculate lift.</p>
<p>The rudder torque is the response to the hydrodynamic pressure on the moving part of the hydrofoil. A perfect spade rudder would be long and thin (think aircraft wing), with its axis approximately 1/3 to 1/2 chord from the leading edge at the centre of lift. Unfortunately with ships the rudder tends to be short and wide and much less efficient, so to achieve the same force, a larger profile area is needed ( F= 1/2 PAV^2) P=fluid density, A = Foil profile area, V=fluid velocity.</p>
<p>Adding a skeg achieves exactly the same result as an aircraft wing with ailerons. The ailerons change the lift characteristic of the foil in two ways - firstly the fluid flow is diverted, changing the relationship of the pressure over the top and bottom surface; secondly the centre of lift moves forward slightly as the aerofoil deflection increases, and therefore increasing effort is needed to operate the aileron until stall is achieved. However the actual effort required with a skeg vs. a free spade rudder is significantly lower since the fixed skeg is part of the foil and carries most of the load.</p>
<p>So the answer to your question depends on the physical geometry required, the fluid flow rate, fluid density, and the aerofoil shape itself. </p>
<p>Some of the low speed Selig or NACA series foils may be suitable - eg NACA 0015 with the movable portion of the foil (rudder) at 50-60% chord from the LE of the skeg. Obviously a leading edge radius on the rudder and a trailing edge radius on the skeg are required. The gap between should be minimised.
<a href="https://i.stack.imgur.com/dHzvO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dHzvO.png" alt="NACA0015"></a></p>
<p>Useful resources:-
<a href="http://www.mh-aerotools.de" rel="nofollow noreferrer">http://www.mh-aerotools.de</a>
<a href="http://airfoiltools.com" rel="nofollow noreferrer">http://airfoiltools.com</a></p>
|
12246
|
Hydrofoil for skeg-hung rudder with forward centre-of-effort
|
2016-11-10T22:46:45.373
|
<p>Large steam engines seem to get pretty good efficiency, but there are other options such as combustion turbines.</p>
<p>Ignoring all other considerations, what methodology can extract the most energy out of a given amount of fuel?</p>
|
|mechanical-engineering|thermodynamics|
|
<p>We need to go directly from chemical energy to electrical energy. Electrical energy can be attained by accelerating electrically charged particles through a coil or magnetic field, which causes electromotive force (voltage) and electrical current through a load to do work to neutralize the voltage.
The chemical reaction of fuel with oxygen is known as oxidation reduction and is exothermic. That means that the end products of the reaction are more stable in terms of the resulting molecule's outermost electron sub-orbitals and pairing of electron spin defined by Pauli's exclusion principle and quantum mechanics.</p>
<p>Since the end product is more stable, energy is released (exothermic). This energy is manifested as photon emission as the electrons settle into the lower energy level orbitals, and as increase of the average kinetic energy of the molecules (heat) in this exhaust.</p>
<p>According to Boyle's law, the increased temperature of the exhaust gas causes expansion of said gas to fill a greater volume (less dense). This thermal expansion of the exhaust gas is what drives the turbines in a gas turbine electric power generation setup, jet engine, rocket engine, and what pushes the pistons inside the cylinders in an internal combustion engine.</p>
<p>Now, the trick to converting directly to electrical energy is to ionize the molecules of the exhaust gas at the point of combustion and keep these molecules ionized as they accelerate through a magnetic field producing electrical energy.</p>
<p>The efficiency of this setup would be determined by how much of the exhaust gas can be ionized and how much of the thermal energy of this ionized exhaust could be extracted. As the charged gas goes through the magnetic field it would slow down (cool) and/or possibly lose charge. The greater the temperature difference between the point of combustion and the end of the magnetic field pipe, the greater the efficiency.</p>
<p>We are converting the thermal kinetic energy of the ions directly to electrical energy.</p>
<p>Or, we can extend (stretch) the entire flame, or ionized process of combustion through the entire insulated tube which would be surrounded by magnetic field.</p>
<p>A straight linear tube would avoid x-ray photon emission caused by rotational acceleration if the tube were curved, but this x-ray emission could be useful in keeping the gas ionized. So a circular or spiraling setup may also work.</p>
<p>Also, a pulsed gas setup may work better than a continual stream. The frequency of pulsation could be fine tuned to the physics of the system for optimal efficiency.</p>
|
12249
|
What is the most efficient way to convert a fuel into electrical energy?
|
2016-11-11T05:26:19.467
|
<p>I recently had an idea about the design of an internal combustion engine. Normally, the spark must be timed using some kind of circuit, but if there was a "spike" on the piston:</p>
<p><a href="https://i.stack.imgur.com/c7FBS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c7FBS.png" alt="enter image description here"></a></p>
<p>then would there be no need for a specialized ignition system?</p>
<p>Of course, there would have to be some kind of insulation between the spike and the casing, and between the spike and the piston.</p>
|
|mechanical-engineering|electrical-engineering|combustion|
|
<p>Well, you'd still need something to generate the high voltage for the spark, so there isn't much overall simplification. A magneto driven by points is about as simple as it gets.</p>
<p>But more importantly, the spark does not always occur at TDC — in fact, in most operating regimes for most engines, it <em>never</em> does. In order to get the best efficiency, the ignition timing must be variable relative to the piston position.</p>
<p>Normally, at idle, ignition occurs a few degrees before TDC. But as the speed goes up, or under hard acceleration, the ignition might be advanced by as much as 40°-50°. Back when we used mechanical points for ignition timing, these were known as "centrifugal advance" and "vacuum advance", respectively.</p>
|
12255
|
IC engine where the piston is the spark plug
|
2016-11-12T21:14:36.873
|
<p>Why does the ballast of train track need good drainage?</p>
<p>I read something about preventing "fouling" of the ballast, but don't know what that means or why it would be structurally bad.</p>
<p>Could foundational support play a role, where soggy ground would give way under the track?</p>
<p>I'm also trying to understand why ballastless track apparently doesn't need good drainage. That continuous slab of concrete surely is not permeable to water.</p>
|
|rail|transportation|foundations|
|
<p>It's also important to drain the ballast and slab track in particular to limit the propagation of stray currents and prevent corrosion of the slab rebars or the track material . This applies to all electrified tracks.</p>
|
12272
|
For train track, why does the ballast need "good drainage"?
|
2016-11-13T14:34:53.690
|
<p>I have encounter this notion: "Dimensionless Thickness". Does anyone have any idea how to interpret this? One can find such a term <a href="https://www.sharcnet.ca/Software/Ansys/17.0/en-us/help/flu_am/th_bat_MSMD_abuse.html" rel="nofollow noreferrer">here</a>.</p>
<p>My interpretation is that the name comes from the fact that the thickness alone doesn't mean too much, but it means a lot if it is related to a reference number (e.g. $ratio = t/t_{reference}$, where t is the dimensionless thickness). </p>
<p>If this is true, can anyone provide me with an example of such a term in the fluid-dynamics, mechanics or thermodynamics field?</p>
|
|mechanical-engineering|chemical-engineering|building-physics|
|
<p>What matters is not the dimensional value of the thickness but rather the ratio with a reference thickness like you stated.</p>
<p>Examples I can think of are when dealing with turbulence:
$U^+$ and $y^+$ which are the non-dimensional velocity and distance from the wall, respectively, when discussing the Law of the Wall in Turbulence.</p>
|
12279
|
Dimensionless Thickness
|
2016-11-14T09:25:05.447
|
<p>I'm trying to find the best way to move a heavy weight around. Probably the best way is to put it on wheels but I don't know which physics concepts it involves so I can make a better research.</p>
<p>I have found about rolling force and initial force but it's still unclear for me what can I do to decrease the force needed. What I found but not sure:</p>
<ul>
<li>Increase diameter of the wheels</li>
<li>Wheels/floor made of low rolling resistance coefficient</li>
<li>Decrease the weight (can't be done)</li>
<li>Good bearings</li>
</ul>
<p>Is this correct? What about the number of wheels, moment of inertia? Am I missing something else?</p>
|
|mechanical-engineering|wheels|
|
<p>As you say wheels are usually the first option considered. </p>
<p>Probably the simplest option is to look at an industrial wheel/castor supplier. For example <a href="http://www.rosscastors.co.uk/?gclid=CIDLnLajqNACFQ06Gwody4sJTw" rel="nofollow noreferrer">this site</a> has a huge range of wheels castors and other parts for moving and handling. </p>
<p>Wheels are typiacly rated with a maximum load, note that if you use 4 castors it is usual to make sure that they are rated so that 3 are adequately rated for the total load ie if the load is 1000kg you need castors rated to at least 333kg. </p>
<p>Castors are generlaly pretty good for reasonably smooth surfaces like a decent concrete floor. Soft or uneven surfaces may require pneumatic wheels which tend to have significantly larger diameters. </p>
<p>Castors have the advantages of being self contained, vesatile and compact and they can often be fixed directly to th estructure you want to move but only work well on reasonably smooth, hard surfaces. </p>
<p>Typically harder wheels give less rolling resistance on smooth surfaces but are more affected by uneven floors. </p>
<p>Inertia is a fundamental property of matter and there is nothing you can do to decrease it. But when moving things by hand the ability to control heavy loads on wheels becomes a problem before they are too massive to get moving. </p>
|
12289
|
Force required to move object on wheels
|
2016-11-15T00:12:34.990
|
<p>I need to build a device that has motorized a door with a hinge on top, similar to cat doors:</p>
<p><img src="https://i.stack.imgur.com/ruIaJs.jpg" alt=""></p>
<p>Between these options, which one makes the most sense:
- use an actuator on the side to push the door up
- put a pulley on the door hinge and drive it with a stepper motor through a belt
- put an anchor above the door, on the same plane, and pull a string on a reel with a stepper motor
- or, any other idea :)</p>
<p>What makes the most sense from a mechanical point of view? it's for a small device where the door is roughly 7x7" and weights about 0.6LBS.</p>
<p>The actuator seems the simplest so far, but before I order parts I would like to have opinions from people that understand this stuff :) </p>
|
|mechanical-engineering|
|
<p>All of your suggested methods seem workable.</p>
<p>To me it seems that the major differences are:</p>
<ol>
<li><p>Whether it resists opening the door by external force. A linear actuator would, while a string wouldn't. It depends on your use case which is more desirable, or whether it matters at all.</p></li>
<li><p>Amount of force produced. A linear actuator is the strongest, but if there is a possibility of someone being stuck in between the door you might actually prefer a pulley that can slip without causing damage.</p></li>
<li><p>How to find the zero position. After power is cut out and restored, your device must somehow know what is the current position of the door. Typically this would be done with a microswitch at one of the end positions. This would probably be easiest to place in the closed position for the linear actuator or pulley system, and in the open position with the string system.</p></li>
</ol>
|
12306
|
Automatic door, linear actuator or stepper motor
|
2016-11-16T03:10:44.273
|
<p>When I see electrical generators specified, they usually give the fuel consumption as at "rated load". So,for example, a 5 kilowatt generator might consume 0.6 gallons per hour.</p>
<p>However, what happens if the load is less than 5 kilowatts? Does the generator still run at the fixed rate and any excess electricity is just shorted to ground?</p>
<p>Or does the engine run slower and just generator less energy? How does it know what rate to run at? Does it have to manually adjusted?</p>
<p>If the engine is run at less than its maximum load, what would be a typical loss of efficiency? Can someone provide an example curve that shows how a generator loses efficiency at sub-optimal loads?</p>
|
|generator|
|
<p>There are two types of generator sets of interest: regular generator sets and inverter generator sets.</p>
<p>Regular generator sets create 50 Hz or 60 Hz electricity directly all the time. Thus, the rotation rate of the generator always has to be 3000 RPM or 3600 RPM. However, this doesn't mean the engine is running at 3000 RPM or 3600 RPM. The engine may be connected to the generator by a belt, by a chain or by using gears. Nevertheless, the engine is all the time running at constant speed, whatever it may be. In directly connected generators (no gears, belts or chains), obviously the running speed of the engine has to be 3000 RPM or 3600 RPM.</p>
<p>Because the engine is running at constant speed, it is never "idling" if you don't have any load. The load on the engine decreases, but its speed doesn't. If it's a gasoline engine, the throttle plate will be nearly closed with no load, creating lots of pumping losses. Friction losses also don't decrease. Therefore, the fuel consumption will be reduced only very little if you disconnect all load. Diesel generators would be slightly better because they have no pumping losses, but they would still suffer from the constant friction in the engine. I don't have any figures for regular generator sets, but expect them to be poor below the rated load.</p>
<p>Inverter generator sets are different. They create high-voltage direct current, and then use pulse width modulation to synthesize sine wave, or for some early or really cheap (not common anymore nowadays) generator sets, modified square wave with three levels: positive, zero, negative.</p>
<p>Because the frequency of the output sine wave is synthesized, the engine may be running at any speed provided that the engine can supply enough power. Usually there's an "economy switch". Some cheap Chinese generator sets may say in the manual that "economy switch" must be off for high loads, better generator sets can even provide full load with "economy switch" on. What the "economy switch" does is it allow free RPMs. With "economy switch" on, RPM is typically variable, very low at no power output, very high at full power output. With "economy switch" off, RPM is always at the maximum power RPM. In better generators (not those cheap Chinese ones), you only need the "economy switch" off if you are planning to start a really large tool or appliance such as a large circular saw, that can require a high inrush current when starting, creating a momentary high load. Typically these inverter generator sets are designed to be lightweight, able to be carried with one hand, so the flywheel is lightweight, and given that the engine is single cylinder, it can stall if there's a huge inrush current. When you turn the "economy switch" off, you increase the RPM of the engine, allowing starting huge tools or appliances. Then you can immediately turn the switch on after the inrush current is gone and the engine stays at optimal RPM.</p>
<p>Even inverter generator sets that run on gasoline suffer from pumping losses. At very low load, the throttle plate is nearly closed, creating an intake vacuum that causes those pumping losses. Diesel generators would be better but heavier and more expensive. The main benefits of inverter generators are that (1) if RPM decreases, friction losses decrease, and (2) if RPM decreases, throttle plate can be opened more for same power output, which reduces but does not eliminate pumping losses. Also noise reduces if engine speed reduces.</p>
<p>Here's an example of fuel consumption of Honda EU10i (1000 watt inverter generator set): <a href="https://hsaoy.com/wp-content/uploads/PAKM_Honda_EU10i.pdf" rel="nofollow noreferrer">https://hsaoy.com/wp-content/uploads/PAKM_Honda_EU10i.pdf</a></p>
<p>In case the link goes stale, here are the figures:</p>
<pre><code>0 W, 0.25 l/h
215 W, 0.29 l/h, 1.35 l/kWh
430 W, 0.38 l/h, 0.88 l/kWh
630 W, 0.47 l/h, 0.74 l/kWh
829 W, 0.59 l/h, 0.71 l/kWh
1029 W, 0.77 l/h, 0.75 l/kWh
</code></pre>
<p>So you can see that at full load, it's consuming about 3x the gasoline it would be consuming at no load. That's hardly optimal. A car cruising at 120 km/h is consuming maybe 7 liters / 100 km, or 8.4 liters per hour whereas a car idling would be consuming 0.5 liters per hour, 16.8x difference. And the "cruising at 120 km/h" wouldn't even be full load for the car. Whereas for these small inverter generators, full load consumes 3x the idle consumption.</p>
<p>Gasoline contains about 32 MJ/l or 8.89 kWh/l, so the best efficiency this generator set (Honda EU10i) can attain is below 16%. Typical efficiency at low load is maybe 10%. As you can see, most of the energy goes into keeping the engine running, and very little goes into the produced electrical power output, except at high loads where produced electrical power output exceeds the constant consumption.</p>
<p>A larger generator would be less efficient at same load. At 386W average load, my Champion 92001i (1900 watt unit) consumes 0.44 liters per hour of small engine gasoline (which would correspond to 0.42 liters per hour of ethanol-less regular gasoline -- the small engine gasoline has low energy density). The EU10i at same load would consume about 0.36 liters per hour (interpolated from the figures), not sure if the EU10i test was done with ethanol-less regular gasoline, with E10 gasoline or with small engine gasoline.</p>
|
12323
|
Do generators run slower if the load is lower?
|
2016-11-16T17:33:05.040
|
<p>If an air pump needs to able to generate a particular pressure at a particular flow rate. </p>
<p>How do we size the electrical motor for this particular scenario? The electric motor would be coupled to the pump shaft in this case.</p>
<p>I mean how do I calculate the mechanical power the motor shaft has to deliver to achieve this? </p>
<p>Does the pump technology have any bearing on this I mean diaphragm, piston etc.</p>
<p>Lets take an example pressure of 1 psi at a flow rate of 10 liters / min.
I need to know the wattage of the motor that would be a suitable fit for this specification.</p>
|
|pressure|pumps|power|
|
<p>Generally speaking, for fluid systems the mechanical power can be computed as the product of pressure and volumetric flow rate:</p>
<p>$\dot{E} = P\dot{V}$</p>
<p>where $\dot{E}$ is the power, $P$ is the pressure and $\dot{V}$ is the flow rate. Therefore, for your application (with unit conversions to get the output in watts):</p>
<p>$\dot{E} = (1 \text{ psi})(\frac{6894.76 \text{ Pa}}{1 \text{ psi}})(10 \text{ L/min})(0.001 \text{ m}^3/\text{L})(\frac{1 \text{ min}}{60 \text{ s}}) \approx 1.15 \text{ W}$</p>
<p>However, the required flow and pressure are not the only considerations. You need to consider all the sources of friction and other energy losses in the system: friction in the motor bearings, efficiency losses when converting mechanical energy in the impeller to fluid flow in the system, etc. Therefore you would need an electrical motor with more than just 1.15 W of power. The energy losses from the impeller to the fluid flow would be difficult to estimate if you're building your own pump.</p>
<p>Also consider that an electric motor cannot output 1.15 W for any combination of speed and torque, so you need to check the motor's spec sheet to make sure that it can provide the power you need at the desired motor speed.</p>
|
12342
|
Relationship between output mechanical power and max pressure for a pump?
|
2016-11-17T04:31:33.917
|
<p>I'm designing a gear system which is 1 motor(drive gear) will drive the last gear(driven gear), and there will be a few gear between the motor and output gear. I understand that the gear ratio between motor and output gear didn't change even I add how many gear in between, and even I change the size of gears in between. As long as I keep the input and output.</p>
<p>Here the link reference for above explanation <a href="http://www.wikihow.com/Determine-Gear-Ratio" rel="nofollow noreferrer">http://www.wikihow.com/Determine-Gear-Ratio</a></p>
<p>My question is, what happen if I add cluster gear in between, does it change the ratio between drive gear and driven gear? As what I understand, it will not change the gear ratio, but a few engineer's friend told me different thing. I cannot found solid explanation on this.
Please refer attached image for what I mean.</p>
<p><a href="https://i.stack.imgur.com/zUjhh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zUjhh.png" alt="enter image description here"></a></p>
<p>Thank you.</p>
|
|motors|gears|
|
<p>Having a sequence of gears connecting the driving and driven gears that all mesh in the same plane will produce a gear ratio independent of the intermediate gears: the gear ratio will be the number of teeth on the driven gear divided by the number of teeth on the driver gear. The following image shows this type of arrangement (not all teeth are shown):</p>
<p><a href="https://i.stack.imgur.com/xyrUd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xyrUd.jpg" alt="enter image description here"></a></p>
<p>Note that this is also the case for arrangement 2 because, even though there exists a gear out of plane, all of the gears involved in transmitting rotation from driver to driven all lie in plane.</p>
<p>However, whenever you have rotation transmitted from one gear to a gear of a different size via a shaft, as is the case for the cluster gear in arrangement 1, the gear ratio does depend on the intermediate gears.</p>
<p><strong>"IN-PLANE" GEAR TRAIN</strong></p>
<p>The reason why the gear ratio is independent of intermediate gears for a simple chain is because all the meshing gears have the same circumferential speed (to be more precise, the tangential speed of a gear at its pitch circle): any two meshing gears must have the same circumference speed, and this extends to all the gears because they all lie in plane.</p>
<p><a href="https://i.stack.imgur.com/H2pFK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H2pFK.jpg" alt="enter image description here"></a></p>
<p>In the above diagram, I use the following notation: $V$ is the circumferential speed, $\omega_D$, $\omega_d$ are the angular velocities and $r_D$, $r_d$ are the pitch circle radii for the driving (D) and driven (d) gears respectively.</p>
<p>Given $V$ is the same everywhere in an in-plane gear train, it can be shown that, according to kinematics:</p>
<p>$$V = \omega_D r_D = \omega_d r_d$$</p>
<p>And so the (reduction) gear ratio is:</p>
<p>$$G = \frac{\omega_D}{\omega_d} = \frac{r_d}{r_D} \implies G = \frac{N_d}{N_D}$$</p>
<p>$N_D$ and $N_d$ are the number of teeth on the driving and driven gears respectively.</p>
<p>And so it can be seen that the gear ratio is indeed independent of the intermediate gears.</p>
<p><strong>COMPOUND GEAR TRAIN</strong></p>
<p>For gear trains like in arrangement 1, where at least one pair of gears involved in transmission motion are linked by a shaft, the circumferential speed is not the same everywhere:</p>
<p><a href="https://i.stack.imgur.com/8kgAm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8kgAm.jpg" alt="enter image description here"></a></p>
<p>In the above, by kinematics:</p>
<p>$$\omega = \frac{V_1}{r_1} = \frac{V_2}{r_2} \rightarrow r_1 \ne r_2 \implies V_1 \ne V_2$$</p>
<p>Instead, to determine the gear ratio of such a system, it helps to consider gear ratios of sub-systems consisting of only in-plane gear trains:</p>
<p><a href="https://i.stack.imgur.com/i5EQR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i5EQR.jpg" alt="enter image description here"></a></p>
<p>In the above diagram, there are two sub-systems of in-plane gears: the lower set and the upper set. Consider the lower set of gears having its own gear ratio, noting that all the gears in this set are in-plane:</p>
<p>$$G_1=\frac{\omega_D}{\omega_1}=\frac{N_1}{N_D}$$</p>
<p>Similarly, for the upper sub-system of gears:</p>
<p>$$G_2=\frac{\omega_1}{\omega_d}=\frac{N_d}{N_2}$$</p>
<p>Therefore, the overall gear ratio can be determined:</p>
<p>$$G = \frac{\omega_D}{\omega_d}=\frac{\omega_D}{\omega_1} \frac{\omega_1}{\omega_d} = G_1 G_2$$</p>
<p>Therefore, it can be shown that for a compound (out-of-plane) gear train, the gear ratio is equal to the gear ratios of in-plane sub-systems all multiplied together.</p>
|
12345
|
Does Cluster Gear Change The Gear Ratio
|
2016-11-17T08:38:46.820
|
<p>Where can I find (cite-able sources) properties (such as specific heat at constant volume, thermal conductivity, viscosity etc...) of air at different temperatures (Range of 200K to 2000 K)?</p>
|
|standards|
|
<p>Cengel's Book: Yunus A. Cengel;John M. Cimbala;Robert H. Turner, Fundamentals of Thermal-Fluid Sciences. 4th. McGraw-Hill, 2012</p>
<p>Here is an extract of the property tables: <a href="http://www.kostic.niu.edu/350/_350-posted/350Chengel7th/Appendix1Udated.pdf" rel="nofollow noreferrer">http://www.kostic.niu.edu/350/_350-posted/350Chengel7th/Appendix1Udated.pdf</a> (for air, see A-2, A-16 and A-17)</p>
<p>There is also: Yunus Cengel and Afshin Ghajar, Heat and Mass Transfer, Fundamentals and Applications, 4th McGraw-Hill, 2011 (for air, see A-15)</p>
<p>They have all the properties of air at 1 atm ($\rho , C_p, k, \alpha , \mu , \nu, Pr$) from -150 $^{\circ}$C to 2000 $^{\circ}$C. See <a href="https://www.scribd.com/doc/30589475/Heat-Transfer-Appendix-tables" rel="nofollow noreferrer">here</a>.</p>
|
12348
|
Standard properties of air
|
2016-11-17T08:44:32.970
|
<p>I have some questions about the enthalpies between two points during an isentropic process in a nozzle. By doing the energy balance between the two points, one gets:</p>
<p>$h_1 + \frac{v_1^2}{2}+gz_1=h_2+\frac{v_2^2}{2}+gz_2$</p>
<p>If one neglects the potential energy and assuming that the velocity at the inlet is 0, one ends up with this:</p>
<p>$h_1 =h_2+\frac{v_2^2}{2}$</p>
<p>This is the part that raises some questions. If the process is isentropic, it means that the entropy doesn't change and therefore there is no heat added (or vice-versa) to the flow. But wouldn't that make $h_1 = h_2$? </p>
<p>My guess is that because $h = u + Pv$, the Pv term is the one that makes them different, considering that the flow energy (Pv) is different. But this would make $u_1 = u_2$. Is this true?</p>
<p>And if this is true, why don't we just use the following relation:</p>
<p>$P_1v_1=P_2v_2+\frac{v_2^2}{2}$</p>
<p>At this point, one can't really define $Cp$ or $k=Cp/Cv$, which would lead to the well-known isentropic flow equations. Am I getting something wrong? Can anyone help me to understand this? Thank you!</p>
|
|fluid-mechanics|thermodynamics|airflow|compressible-flow|
|
<p>Constant entropy doesn't mean constant temperature, the gas flow through nozzle is subject to change in pressure and velocity (continuity equation) due to the variation in cross section, and being compressible the change in pressure will be reflected as change in temperature and density.</p>
<p>That being said, since the flow is isentropic and there will be no energy dissipation due to friction and by applying the first law of thermodynamics to the nozzle you will (naturally!) find that the sum of energy across the nozzle is constant, meaning that the stagnation enthalpy ($h_1$) will equal the sum of enthalpy and kinetic energy at every other point in the nozzle (any increase in kinetic energy will be reflected as decrease in enthalpy and vice versa).</p>
<p>So, to answer your question: $h_1 \neq h_2$ and also the sensible temperature $T$ reflected in internal energy is not necessarily the same for the two point in comparison, meaning that also $u_1 \neq u_2$, you have to look at the big picture where the change in cross section will affect the velocity, pressure, temperature and density (that's why the analysis of compressible flow is far more complex than incompressible flow).</p>
|
12349
|
Shouldn't enthalpies be equal in isentropic process?
|
2016-11-17T23:49:05.767
|
<p>I'm intrigued by <a href="https://www.youtube.com/watch?v=TsaES--OTzM" rel="nofollow noreferrer">this video</a> in which the roof camera rotates continuously. One reason could be to capture 360 degrees around, but then again two cameras or more could do the same job capturing different parts of the landscape. It seems to me a waste of energy in making the camera rotate, and also having mobile parts make it more vulnerable to breaking. I don't see why not using a system of many static cameras.</p>
|
|control-engineering|car|
|
<p>That's a <a href="https://en.wikipedia.org/wiki/Lidar" rel="nofollow noreferrer">lidar</a> sensor, which is making a continuous 360 degree sweep, rather than a camera which has a field of view as such. Rotating mirrors are a fundamental part of how the system works. </p>
<p>Note that in the video the system represents traffic as fairly crude boxes rather than rendered images, so it may be looking specifically at reflectors, which also helps it to identify signs and traffic cones. </p>
<p>Also, they are talking very much about the software and control systems, so it seems likely that the sensor rig in the video is chosen to be convenient for testing, rather than a fully developed and refined consumer version. </p>
<p>In practice, a fully autonomous car may well need a variety of different sensors to guarantee redundancy and effectiveness on a variety of conditions, but in good weather with a professional human driver supervising the vehicle, a bolt-on lidar sensor is a convenient way to fine tune the software. </p>
|
12366
|
Why does "Google self-driving car"'s roof camera rotate?
|
2016-11-18T22:18:43.940
|
<ol>
<li><p>I sometimes design my retaining walls with longer toes and heels, this is because in my FEA program, when I pull my footing under my wall towards the toe, I get lower pressures on my soil (springs) that when I pull it towards the heel (even with the added benefit of the earth on top). This tends to happen when I don't have too much counter balancing soil on my heel. Anyone else get these results? I've never seen any long toe retaining walls in the literature.</p></li>
<li><p>When calulating over turning moment by hand, all of the literature I've seen takes the point of rotation at the toe of the wall. If I look at my FEA design, the wall seems to be rotating much more around the heel. Is it justifiable to calculate the overturning moment at the toe? This seems to be the point that's most favorable for design, but is it realistic?</p></li>
</ol>
<p><a href="https://i.stack.imgur.com/KgCPY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KgCPY.png" alt="enter image description here"></a></p>
|
|structural-engineering|geotechnical-engineering|
|
<p>I have designed plenty of (probably several dozen) retaining walls with long toes in situations with low allowable bearing capacities. Usually designed with RetainPro (which is not FEA), but also using "by hand" calcs as well. So your results are not unusual. </p>
<p>With regards to rotation about the toe or heel: overturning and rotation are two different things. You actually want the wall to rotate about the heel just a bit so you can transition from at-rest to active soil conditions, and relieve the locked-in horizontal soil load (from compaction behind the wall). </p>
<p>However, if the wall is going to catastrophically fail, which is what an OTM check is looking at, it WILL rotate about the toe eventually. At the point in time it fails, such a wall will probably have initially rotated around the heel, but the overturning failure will be about the toe. Given the presence of the soil underneath constraining movement into the soil mass, how could it not? But the walls you are analyzing don't seem to be in danger of catastrophically failing (or your FEA would not converge). </p>
<p>Think of it this way: in order for the wall to fail due to the applied overturning moment about the heel, the footing toe would have to rotate INTO the soil. This cannot happen unless there is a slope failure, but that is a separate discussion topic. And that failure mode is not being analyzed in your FEM- that is, unless it's a geotechnical analysis package and you have some kind of decent soil modeling design parameters, which nobody ever has. </p>
<p>Try this experiment in your FEM: put an X fixity at the footing toe (so it cannot slide). Then incrementally ramp up the horizontal load only (no corresponding increase in vertical load). The structure will eventually rotate about the toe. Checking this kind of load condition- high horizontal loading (usually due to live load surcharge BEHIND the wall heel) with high resistance to sliding- is the purpose of the OTM check. Under such a loading condition, it would make no sense to check the OTM about the heel. The wall simply will not fail that way unless there is a localized bearing capacity failure. </p>
|
12381
|
Postion of overturning moment for retaining wall
|
2016-11-20T23:42:25.383
|
<p>I was given the following equations and asked to design a controller for $u=-Kx$ using a pole placement method with the closed loop system having the damping and settling time $T_s$ given.</p>
<p>Am I supposed to use the $\dot x_1$ and $\dot x_2$ equations or are they totally unrelevant?</p>
<p>Also, which pole placement method is easier for the case, root locus or Bode and Nyquist plots?</p>
<p>$$\begin{gather}
\dot x_1 = x_2 \\
\dot x_2 = -x_1 +\dfrac{1}{6}x_1^5-x_2+u \\
u = -Kx \quad \zeta=1.02 \quad T_s = 0.40
\end{gather}$$</p>
|
|control-engineering|control-theory|stability|
|
<p><strong>Step 1</strong></p>
<p>The first thing to do is to determine the desired poles. </p>
<p>The natural frequency can be computed using $\omega =\frac{4}{\zeta T_s}$. This is a rule-of-thumb calculation for underdamped systems. The system here is slightly overdamped and is nonlinear as well. If the desired settling time is not obtained in the end, we have to come back and increase the constant 4 slightly. The design procedure is typically iterative. So we start with $\omega =9.80392$</p>
<p>Then the characteristic equation can be computed as $s^2+2 \zeta s \omega +\omega ^2$, which after substituting values gives $s^2+20 s+96.1169$ and has roots $-11.9706$ and $-8.02944$</p>
<p><strong>Step 2</strong></p>
<p>Put the system in a linear from $$\dot{x}=\text{A}.x+\text{B}.v$$ where $$x=\left(
\begin{array}{cc}
x_1 & x_2 \\
\end{array}
\right)^T \ \ \ v=u+\frac{x_1^5}{6}$$
$$A=\left(
\begin{array}{cc}
0 & 1 \\
-1 & -1 \\
\end{array}
\right) \ \ \ \ \ B=\left(
\begin{array}{c}
0 \\
1 \\
\end{array}
\right)$$</p>
<p><strong>Step 3</strong>
Do the pole-placement design which gives $v=-k.x$ using Ackerman's formula.
$$k=\left(
\begin{array}{cc}
0 & 1 \\
\end{array}
\right).
\left(
\begin{array}{cc}
\text{B} & \text{A}.\text{B} \\
\end{array}
\right)^{-1}.
(\text{A}^2+20 \text{A}+96.1169I)$$</p>
<p>Substituting values, we get
$$k=\left(
\begin{array}{cc}
95.1169 & 19 \\
\end{array}
\right)$$</p>
<p><strong>Step 4</strong></p>
<p>Do the back transformation to get the value for $u$.
$$u+\frac{x_1^5}{6}=-95.1169 x_1-19 x_2$$
$$u=-95.1169 x_1-19 x_2-\frac{x_1^5}{6}$$</p>
<p><strong>Step 5</strong></p>
<p>Verification. We have to see if the design has met the requirements. (These simulations were done in Mathematica. The above calculations could also have been done there. I went through them manually above to explain things.) From the plot we see that the settling time constraint has been satisfied. </p>
<p><a href="https://i.stack.imgur.com/Y3MG7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y3MG7.png" alt="enter image description here"></a></p>
|
12408
|
Controller design with pole placement method with given damping and settling time
|
2016-11-21T13:13:32.990
|
<p>I need to buy a particular brand and model of stepper motor and I have found the product <a href="http://motion.schneider-electric.com/products/mdrive/mdrive_plus_motion_control.php?nema=17" rel="nofollow noreferrer">on the manufacturer's web catalog</a>, but with no price listed.</p>
<ol>
<li>How can I find the price of device? I tried writing to the manufacturer but they haven't responded. </li>
<li>Why has the manufacturer hidden the price? </li>
</ol>
|
|mechanical-engineering|equipment-selection|
|
<p>In industry, an equipment manufacturer hardly ever sells directly to the end user. They supply distributors in bulk at a negotiated price for each shipment of however many units the distributor orders. The price-per-unit varies accordingly, so there's no point in listing it the way you would for retail products.</p>
<p>If you only need a single unit of a product, generally the manufacturer is not going to want to do business with you. They produce many units at a time, as required to supply their distributor(s), in order to use their facilities and personnel efficiently.</p>
<p>What you need to do is find a distributor or wholesaler who carries that manufacturer's products and is willing to sell directly to you (retail business). <a href="https://en.wikipedia.org/wiki/W._W._Grainger" rel="nofollow noreferrer">Grainger</a> is one example of a company that does this, though you may have to work for a company that has a business account with them in order to just walk in and order a single item.</p>
<p>You can usually find the contact information for a local or regional distributor on the company's web site or in their catalog, provided they do business in your area. Sometimes, especially when dealing with a foreign supplier, they won't have an existing business relationship with anyone in your region, and then you may need to look for an alternative product or work with a foreign distributor as well, which can be a real headache for just one piece of equipment.</p>
|
12415
|
How can I find the price of a product if the manufacturer doesn't show it on their website?
|
2016-11-21T15:39:34.643
|
<p>I'm building a hovercraft, which transfer function i have to find for further control. The hovercraft can be found in the following <a href="https://i.stack.imgur.com/g9xVG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g9xVG.png" alt="enter image description here"></a></p>
<p>As you can see, the hovercraft has 2 motors for direction. I have been given an advice to apply a step to the system, via the direction motors, and determine the transfer function, from the stepresponse. For this several questions comes to my mind:</p>
<p>A step on the motor, meaning for example 5V on the motors?</p>
<p>What should the ouput of the transfer function represent? (should i monitor velocity, acceleration, position after x seconds?)</p>
<p>Best regards</p>
|
|transfer-function|
|
<p>If you want to determine the transfer function of a system by applying a step response, you'll need to decide in advance what kind of dynamic model you expect your system to have. If you're looking for a transfer function, that means you've already assumed your system is linear, so start considering linear models that would work for your case.</p>
<p>Once you decide on a model, derive its transfer function. Apply a step response and use the response characteristics (such as time constant, natural frequency, steady state value, or any others that are relevant to your model) to solve for the unknown parameters of the model transfer function. </p>
<h1>Example</h1>
<p>Let's say your hovercraft primarily has two forces acting on it: the force from the motors, and the force due to friction/drag. In that case you could approximate the dynamics using a classical linear, single degree-of-freedom model, like so:</p>
<p>$m\dot{v} = K_m u(t) - bv$</p>
<p>where $v$ is the velocity of the hovercraft, $\dot{v}$ is the acceleration, and $u$ is the voltage applied to the motors. The unknown parameters of the system are: $m$ (the mass of the hovercraft), $K_m$ (the conversion between applied motor voltage and thrust force), and $b$ (the drag/friction coefficient). You should be able to identify $m$ by simply weighing your hovercraft. That leaves $K_m$ and $b$ unknown.</p>
<p>You can get the transfer function of the model by applying the Laplace transform:</p>
<p>$sV(s) = K_m U(s) - bV(s)$</p>
<p>$\frac{V(s)}{U(s)} = \frac{K_m}{ms + b} = \frac{K_m/b}{(m/b) s + 1}$</p>
<p>Now, you can identify the different parameters by measuring the velocity of the hovercraft when you apply a step function to the input. The step function can have any value, but you want it to be large enough that the velocity is measurable, and small enough that the hovercraft doesn't speed out of control.</p>
<p>You are looking for two characteristics of the response:</p>
<ol>
<li><p>The steady state velocity $v_{ss}$, which is the maximum speed that the hovercraft reaches for a given input, even if left for an infinite amount of time. Your hovercraft will probably stop accelerating after a few seconds if your input isn't too large, use that velocity as an approximation. Solve for the ratio of unknown parameters using $v_{ss} = (K_m/b)u_0$ if $u_0$ is the magnitude of the step input, maybe 2 V or something.</p></li>
<li><p>The <a href="https://en.wikipedia.org/wiki/Time_constant" rel="nofollow noreferrer">time constant</a> $\tau$ of the response, which is the time it takes for the system to go from 0 to approximately 63.2% of the steady state velocity (see the Wikipedia link for more info). Solve for $b$ using $\tau = m/b$.</p></li>
</ol>
|
12416
|
Transfer function for hovercraft
|
2016-11-21T20:58:16.100
|
<p>Okay, Hi there Engineers! I am a newbie to this site, and to engineering.</p>
<p>So, I wanted to make this post to get all the help I can get. I have been set a task of creating a prototype of the following;</p>
<p>I have been given a device (the device being a <strong>Rubidium Atom Clock/Standard</strong>. Pictures are provided below.)</p>
<p>The fault with this device is that different internal/external temperatures will affect the clock. So what I need to do is; somehow make the interior temperatures a constant, where ever the device is (so if it is in Iceland, the internal temperature must be the same as it will be if it was in the Sahara Desert.)</p>
<p>Oh yeah, and the interior temperature must be within <strong>-20°C</strong> to <strong>+65°C</strong>.</p>
<p>If you need any information before you can help me, please go ahead and ask. Also, as I've stated above, I am new to engineering, so if you are trying to help me, please explain as much as you can so I can understand. And, I won't have advanced tools available, just assume that I have the tools in a workshop.</p>
<p>Bear in mind, I have very little space to work with in the device, and watch out for voltage, I don't want any systems like fans or something to affect the power supply.</p>
<ul>
<li><p><strong>Specifications</strong>
(Not exact to the device I am using but I will get that as soon as possible, they are pretty similar):
<a href="https://i.stack.imgur.com/CbRc6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CbRc6.png" alt="Specifications" /></a></p>
</li>
<li><p><strong>Device Images:</strong>
...That is rather annoying. I can't upload more pictures due to my account not having 10 reputation. If you are looking for the device, just search up "Rubidium Oscillator PRS10B"</p>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/mbp13.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mbp13.jpg" alt="Interior" /></a></p>
|
|electrical-engineering|power-electronics|temperature|
|
<p>You first need to set your operational criteria:</p>
<ul>
<li>What is the minimum environment temperature:</li>
<li>What is the maximum environment temperature:</li>
<li>What are the physical size constraints?</li>
<li>How much additional power can I use to maintain this temperature?</li>
<li>How constant does the temperature need to be? (ex +/-1degC)</li>
<li>Does only a portion of the unit need to have a constant temperature?</li>
</ul>
<p>I would not recommend messing around with cooling if you want the temperature to be very constant. I would operate the unit at the high end above your highest environmental temp. Perhaps something like 40C as your constant temp and use a heating element surrounding the unit to hold the temperature up in cold environments. Heating also takes less power and equipment than cooling.</p>
<p>If your space constraints are not too tight, some high quality insulation would be the easiest start. Closed cell foam or ceramic fiber would work well. Vacuum insulated like a thermos beverage container might be good too look at too.</p>
<p>If you cant increase the form factor you will need to add a uniform heating element around the outside of the unit and control it with microcontroller and thermistor using a pid loop. This will obviously cost some additional power.</p>
<p>Alternatively, most temperature dependent sensors and systems don't attempt to modify their environmental temperature, but instead use a micocontroller to digitally measure the temperature and compensate the value for that temperature.</p>
|
12424
|
Creating a constant internal temperature that is negligible to the external environment
|
2016-11-21T23:16:07.713
|
<p>My workplace has a policy to provide standing desks upon request, but no policy to provide chairs of matching height. (I work for the government...) We can buy our own, or build our own. However they added a stipulation that the chairs must be bought or built with five or more wheels.</p>
<p>Since five is an awkward symmetry to measure and to sit in, I'd probably build my chair with six wheels. Or perhaps eight because I can start with a square and lop off the corners to make an octagon base. Or maybe make the base a perfect circle and line the bottom with innumerable tiny wheels?</p>
<p>This set me to wondering: is there any situation in which a chair with N+1 wheels is strictly less stable than a chair with N wheels?</p>
<p>To make the question interesting and rule out trivial answers, assume that the wheels are more-or-less at the chair's perimeter and more-or-less evenly distributed. Assume the floor is flat.</p>
|
|wheels|stability|
|
<p>Assuming all the wheels are evenly spaced on the same circle, then more wheels is always more stable than less wheels. However, there is diminishing return as the number of wheels gets large.</p>
<p>The metric of stability is how far from the center of the circle the center of mass can be before the chair tips over. The chair is stable whenever the center of mass is inside the polygon formed by all the wheel points. The worst case is in the center of one of the edges, since these are the closest points on the polygon to the center of the circle. In the limit, with infinite number of support points, the minimum distance to instability is the radius.</p>
<p>We can therefore quantify stability as the minimum distance to instability relative to the radius. A value of 1 is the maximum, with infinite support points. After a little trig, it's easy to see that this stability metric is:</p>
<p> S = cos(Π/N)</p>
<p>where N is the number of support points. The stability metrics for values of N up to 20 are:</p>
<pre>
N S
---- ----
2 0.00
3 0.50
4 0.71
5 0.81
6 0.87
7 0.90
8 0.92
9 0.94
10 0.95
11 0.96
12 0.97
13 0.97
14 0.97
15 0.98
16 0.98
17 0.98
18 0.98
19 0.99
20 0.99
</pre>
<p>Office chairs commonly use N=5, which is a tradeoff between being good enough but not too expensive. The extra 7% stability from adding a 6th wheel isn't worth the cost. Or, put another way, you can achieve the same stability as 6 wheels by using 5 wheels but growing the base another 7% outward.</p>
|
12427
|
Is a chair with N+1 wheels ever less stable than a chair with N wheels?
|
2016-11-22T02:39:33.900
|
<p>I'm designing a 5 axis CNC, and preloading my lead screws has never been an issue for me, but for the first time, I find myself needing to preload a worm gear. How is this done, and what's the best way to do it? Thanks. </p>
|
|gears|
|
<p>For worm gears, the most common method to reduce backlash is by forcing the worm screw tightly against the gear. Many manual rotary tables used on milling machines have this adjustment.</p>
|
12431
|
Preload a worm gear?
|
2016-11-22T14:48:02.600
|
<p>I'm sorry if the question is trivial but for me it is not. I'm wondering what's the principle behind the suction process in a piston pump or in a piston of an internal combustion engine. I mean when the piston goes down, pressure inside lowers and air/water enters. Is it the conservation of momentum?</p>
|
|mechanical-engineering|fluid-mechanics|
|
<p>The particles ( atoms or molecules) of any fluid(liquid or gas) in a container will be in continuous random motion.During that motion they collide on the container walls and their velocity (direction or magnitude or both)changes.By momentum conservation the container wall experiences pressure.So if you move the container walls, the particles travel straight untill they hit the walls in the new position thus they fill the additional volume or at least this how my teacher explained pressure.</p>
|
12442
|
Physical principle behind suction of a piston
|
2016-11-23T12:51:59.090
|
<p>I have noticed that windmills are generally built in empty fields with no trees around, and I've been wondering why... </p>
<p>A windmill is generally way taller than trees and I can imagine that trees don't actually affect the flow (see picture).</p>
<p><a href="https://i.stack.imgur.com/wnaZ7.png"><img src="https://i.stack.imgur.com/wnaZ7.png" alt="enter image description here"></a></p>
<p>But is this the reason why there is nothing around or is there something else that makes them waste so much space around? And is the velocity profile from the first figure a realistic one?</p>
<p>If I search for the theory of external convection, the velocity profile looks like this for flow over a plateau:
<a href="https://i.stack.imgur.com/wPFOW.png"><img src="https://i.stack.imgur.com/wPFOW.png" alt="enter image description here"></a>
In the second case, wouldn't the velocity be higher if there is an obstacle? Or is it because the obstacle is a porous medium that damps the flow? And lastly, is the velocity profile at earth level looking like that? Is there really an origin to it, or is it fully turbulent all the way?</p>
|
|mechanical-engineering|fluid-mechanics|thermodynamics|heat-transfer|
|
<p>There is a non-engineering reason as well: the people most open to placing turbines on their land are farmers who are interested in the extra income.</p>
<p>Turbines are not (yet) placed near built-up areas, which generally leaves large farms, undeveloped lots and nature reserves. Getting a building permit for a commercial enterprise in a nature reserve is close to impossible, and undeveloped lots may be zoned for housing, which would also prohibit the erection of wind turbines.</p>
|
12453
|
Windmills in empty fields. Why no trees?
|
2016-11-23T14:03:28.143
|
<p>Has anyone ever tried running buckling simulations in Nastran where inertia relief is invoked? I am just using a simple plate and I am getting "funny" modes (unrealistically low eigenvalues which correspond to in-plane deformations - no buckling). I went through the documents and they do not seem to suggest that it should be avoided or anything of the sort. In fact the following is stated under <code>PARAM,INREL,-2</code> that <em>"This method leads to indeterminate matrices which are not supported by buckling. If attempted the solution will fail."</em></p>
<p>But i am using <code>PARAM,INREL,-1</code> and the model runs through. The <code>*.f06</code> files looks ok too (<code>OLOAD</code>, <code>epsilon</code>, <code>strain energy</code> and so on..)</p>
<p>Can anybody shed any light or suggestions on how to proceed with this?</p>
<p>Thanks!</p>
<hr>
<p><strong>PS:</strong></p>
<p>I have a plate modeled with CQUAD4s referencing a PCOMP and i am trying to find its eigenvalues. The loads on the plate are a combination of all in plane loads (Nx, Ny, Nxy). Due to the existance of in-plane shear, the setting up of the boundary conditions is not a trivial task and that is the reason inertia relief was considered. Also, with inertia relief in the model and along with the "funny" modes, i also get realistic ones, the eigenvalues of which match those i get with ESAComp and ESDU 81047. <em>So to summarise:</em></p>
<ul>
<li>If i could get the eigenvalues <strong>ESAComp</strong> and <strong>ESDU 81047</strong> gives without inertia relief i would gladly do it</li>
<li>The fact that the first realistic eigenvalue i get with inertia relief in the model matches those from above suggests (imo) that it is somehow doable.</li>
</ul>
|
|finite-element-method|buckling|nastran|
|
<p>We apply a few forces for it. The forces are likely to do two different things. To begin with, they will create a rigid body acceleration in accordance with F=ma. I.e. the middle of mass of this item will accelerate dependent on the amount of their forces. However, since we've got a flexible figure, these forces will also be likely to trigger localized deformations and stresses. We do not care about the stiff body acceleration of the middle of mass. We just care about the neighborhood strain.</p>
<p>We can address a transient energetic solution (i.e. sol 109 at NASTRAN), for both the stiff body acceleration as well as also the local deformation, then afterward subtract the stiff body acceleration to receive only the neighborhood deformation we care for. We're spending lots of CPU resources to calculate something which we're simply going to through out anyhow.</p>
<p>So what we would like to do would be to resolve only a static answer (i.e. sol 101 at NASTRAN). However, there's an issue. We can't invert it.</p>
<p>We can add a bogus support to the construction someplace, just for the interest of constraining it to find an invertible stiffness matrix. However, the reaction forces in this restriction will lead to localized deformations and stresses which aren't real and will wreck the response.</p>
<p>So we've got a issue, we do not wish to address a transient dynamic difficulty since it might take a long time, but a static remedy does not work.</p>
<p>So this is what inertia relief will. We add a restriction. I.e. we create the construction fixed at the border rather than free. Then we put in a distributed body force to the whole arrangement that just balances out the stiff body acceleration. That usually means that the response force at our brand new dummy restriction is zero. However, the response force in the service is zero, therefore it will not have some influence on the regional deformations and stresses.</p>
<p>If you're attempting to model buckling of a free figure that's quickening, then I do not understand why not. The additional fake constraints are just likely to constrain out rigid body movement that should not impact the buckling results.</p>
<p>But in case you've got a constrained structure that's not accelerating, then I really don't see how it is logical. I guess what you're doing is deleting the border conditions entirely (since they're tough to model), therefore fictitiously unconstraining the model, then utilizing inertia relief to place them back into... Inertia relief will attempt and create a body force on your complete structure so as to apply a zero response in the fictitiously generated limitations. However, your arrangement does have actual limitations, so there should be a non-zero response in the actual constraints. So the body force will probably be incorrect.</p>
<p>Not quite certain how NASTRAN manages that body pressure. If it uses it together with all the additional loads in creating the differential stiffness matrix, and then you get the incorrect answer.</p>
<p>So that has been a very long winded answer... bottom line, though it may work, unless there's something particular about how NASTRAN does so I am unaware of, I wouldn't suggest it. Spend a little excess time to acquire the bounds modeled properly.</p>
|
12455
|
Inertia Relief and Buckling SOL105 in Nastran
|
2016-11-24T21:42:50.263
|
<p>I have an overall question that's been bothering me for years. Which is more dangerous for our health, a diesel car or a gasoline car?</p>
<p>This is quite a cliche question and I can find a lot of newspapers reporting that diesel is way more dangerous, but I would like to hear a more engineering/scientific/economic approach to this. Can anyone help me with this? I have a series of other questions that would help me to answer this overall question. Here is goes...</p>
<p>In <a href="https://www.theguardian.com/uk/2013/jan/27/diesel-engine-fumes-worse-petrol" rel="nofollow noreferrer">this article</a>, they write:</p>
<blockquote>
<p>Diesel fumes are significantly more damaging to health than those from petrol engines, according to research which shows that related air pollution contributes to lung disease, heart attacks, asthma and other respiratory problems.</p>
</blockquote>
<p>And they are referencing a nice report written by Department for Energy and Climate Change.</p>
<p>So it is agreed by many that Diesel is more dangerous than gasoline. But at the same time, the diesel engine has a higher efficiency.</p>
<p>Q1: Is the efficiency also reflected in the amount of gases getting out? In other words, an engine with a higher efficiency releases a lower amount of gases?</p>
<p>Then there is a second question, which might sound a bit stupid but,</p>
<p>Q2: Why are people so much for buying Diesel cars (especially in Europe)? They are way more expensive, and yet so popular in Europe.</p>
<p>Q3: Is the lower price of Diesel fuel really making a difference? Or is it the power and reliability?</p>
<p>At the same time, the regulations in US are more strict to Diesel cars.</p>
<p>Q4: Is it because of the emissions or is there another reason for this?</p>
<p>Q5: Is really NOx worse for our health than COx?</p>
<p>Q6: What are the engineering limitations in reducing the NOx's?</p>
<p>I know that some question are not strictly related to engineering, but I want to avoid posting the same questions in other places. However, I would appreciate if you could answer to any of the questions. I would highly appreciate any help in understanding this.</p>
<p>I think this is a very important issue, because most of people (like me) are unaware of these issues.</p>
|
|automotive-engineering|chemical-engineering|gas|petroleum-engineering|diesel|
|
<p>My understanding:</p>
<p>Petrol engines produce far more carbon monoxide. Carbon monoxide is a colourless, odourless and highly toxic gas. This makes running petrol engines indoors far more dangerous than running diesel engines indoors.</p>
<p>Outdoors however carbon monoxide is not so bad. It's lighter than air so it won't tend to persist near ground level. </p>
<p>Diesel engines produce more particulates and NOx, these things aren't going to kill you immediately but chronic exposure is not healthy and since they are heavy they are more likely to remain near ground level where people can breathe them. </p>
|
12474
|
Diesel or Gasoline? Which one is more dangerous to our health?
|
2016-11-24T22:59:15.280
|
<p>I have created a self balancing robot. It uses PID control to control its pitch angle and gets the feedback from gyroscope and accelerometer. But I have not used any sensor to feedback the displacement to the controller. I have just approximated the voltage applying to the motor to the displacement through programming (let suppose 5v to the motor will make the motor to rotate for 1 rev/s and after integrating it , it will give the displacement covered). Through which I was able to maintain the robot at a single position. But now its time write the report and I do not know what to call this type of feedback of displacement.
Thanks</p>
|
|control-engineering|control-theory|pid-control|systems-engineering|feedback-loop|
|
<p>I would call this <a href="https://en.wikipedia.org/wiki/Dead_reckoning" rel="nofollow noreferrer">dead reckoning</a>, because you do not have any actual measurement of the position, but are just calculating it based on the estimated speed of the motor.</p>
|
12475
|
What is the name of my feedback procedure in control system that I am using?
|
2016-11-24T23:49:21.307
|
<p>How large of an area of the earth's surface would we need to cover in mirrors to negate current levels of global warming?</p>
<p>Let's assume that the reflective component of the mirror is made of aluminium foil and that it is kept clean.</p>
<p>And let's also forget about ocean acidification. I'm only really interested in negating the retained heat component of atmospheric carbon.</p>
<p>This question was initially asked <a href="https://physics.stackexchange.com/questions/292765/how-large-of-an-area-of-the-earths-surface-would-we-need-to-cover-in-mirrors-to">here</a> on physics.stackexchange but was rejected because it "was an engineering question".</p>
|
|materials|temperature|environmental-engineering|
|
<p>The idea is not new, but using mirrors on the Earth's surface is a particularly inefficient way to go about it.</p>
<p>A <a href="https://en.wikipedia.org/wiki/Space_sunshade" rel="nofollow noreferrer">Lagrange-point sunshade</a> is actually much more practical, and the thermal analysis has been worked out in some detail. You basically need to block about 2% of the sun's light currently reaching the Earth.</p>
<p>Beware of unintended consequences!</p>
<hr />
<p><strong>Update:</strong> The topic of Earth's heat-flow balance and resulting surface temperature is surprisingly complex. A <a href="https://www.youtube.com/watch?v=oqu5DjzOBF8&t=23s" rel="nofollow noreferrer">video by Sabine Hossenfelder</a> is a good start at "peeling the onion".</p>
|
12477
|
How large of an area of the earth's surface would we need to cover in mirrors to negate global warming?
|
2016-11-25T12:07:48.617
|
<p>Suppose I have a liquid in a container, the type of liquid is such that it expands when its temperature rises. The contrainer is marked with scale to measure the height of liquid expanded. There is a controller that maintains the temprature at it desired value. Temprature sensor feedback the temprature value to generate error signal into the controller. But the main purpose of the system is to maintain the height which does not have any feedback sensor. The only way to find out is through scale reading which is done by human observer. To maintain the height, the observer changes set point temprature. So first, what is the actual name of this kind of system, an open loop system for height control? Second is this block diagram correct in control theory? </p>
<p><a href="https://i.stack.imgur.com/laZpe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/laZpe.png" alt="Control sytem"></a></p>
|
|control-engineering|control-theory|temperature|pid-control|feedback-loop|
|
<p>This system can be considered "closed loop," since the control input is being determined by some sort of feedback loop (even though the mathematical expression of the feedback loop through the operator is unknown). Also, yes the diagram you've shown is valid. I often work with control systems that have a human operator in the control loop, such as robotic teleoperation systems, and they are often shown as some black box operating on a feedback signal.</p>
<p>However, it would be difficult to apply any sort of control theory to your system without some sort of model of how the human operator reacts to the height of the liquid. Often you can make an assumption, like maybe the operator behaves like a proportional controller with a bit of noise (you would need to have solid justifications for any assumptions you make). Either way, in this case if you want to control the height of the fluid with any amount of precision, replace that operator with an ultrasonic rangefinder (or some other sensor) plus some variation on a PID controller.</p>
|
12484
|
What type of control system is this? open loop?
|
2016-11-27T03:32:21.413
|
<p>In the notes , i noticed that the shear stress on the top plane and the front plane will 'meet together' at the edge . I dont understand it why the direction is like that .</p>
<p>Is it possible for both the shear stress to become like this (opposite direction ) ?</p>
<p><a href="https://i.stack.imgur.com/WffLW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WffLW.png" alt="image1"></a></p>
<p><a href="https://i.stack.imgur.com/0GNkA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0GNkA.png" alt="iamge2"></a></p>
<p><a href="https://i.stack.imgur.com/ifJHj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ifJHj.png" alt="image3"></a></p>
<p><a href="https://i.stack.imgur.com/rGUv2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rGUv2.jpg" alt="image4"></a></p>
|
|structural-engineering|
|
<p>Yes, it is. If you draw the shear force arrows on all four sides of a quadrilateral, they point "towards" two diagonally opposite corners, as in the picture in example 7.3(c) and "away from" the other two corners, as in your own picture.</p>
<p>Think about a very small quadrilateral element of the structure, small enough so that the shear and direct stresses can all be considered constant over the element. If the shear force arrows pointed "head to tail" around the sides of the quadrilateral, the forces would produce a moment about the center point, so there would have to be some other forces to counterbalance that moment if the structure was in equilibrium. But uniform <em>direct</em> stresses on the edges of the quadrilateral can't produce a moment about the center. </p>
<p>So, the shear forces must be arranged so that the two forces on one pair of opposite sides produce a moment in one direction, and the two forces on the other pair of sides produce an equal moment in the opposite direction.</p>
|
12515
|
direction of shear stress
|
2016-11-27T09:56:28.537
|
<p>Are these right angle projections correct?</p>
<p><a href="https://postimg.org/image/64l4ibu25/" rel="nofollow noreferrer">https://postimg.org/image/64l4ibu25/</a></p>
<p><a href="https://postimg.org/image/pnpprostp/" rel="nofollow noreferrer">https://postimg.org/image/pnpprostp/</a></p>
|
|mechanical-engineering|
|
<p>Link 6414ibu25 is missing a dashed (hidden) line in the upper left projection on the first image.
The second image in 6414ibu25 appears to be correct</p>
<p>Link pnpprostp - the upper right image is projected from the wrong side.
The lower image in pnpprostp is correct.</p>
|
12520
|
Are these right angle projections correct?
|
2016-11-27T11:13:40.840
|
<p>I want to know how to model and how to use an injection nut in SolidWorks.</p>
<p>The problem starts with me not knowing the correct term in English for them thing I want to model (<em>Einpressmutter</em>).</p>
<p>I'm referring to a metric nut (with metric thread, if course) which exposes a circular, radial notch at one end. This end is to be forcefully inserted into sheet metal with a hole slightly smaller in diameter than the nut's end.</p>
<p><a href="http://www.pemnet.com/fastening-products/nuts-for-sheetmetal/" rel="nofollow noreferrer" title="Examples for those devices">Penn Engineering nuts for sheet metal</a></p>
<p>There is no problem for me in modeling such a part. What I want to have, is like a combination of a part and a forming tool. I.e. I want to drag this part into a closed sheet metal area and the holes for the nut should be created automagically. </p>
<p>Google didn't reveal anything useful. I found only tutorials in creating regular nuts, yawn. The SW tutorials weren't helpful either. But this, again might be caused by using the wrong search term. </p>
<p>I'd appreciate <strike> a link to </strike> a tutorial<strike>s</strike> as well as a better search term or a basic outline of the method.</p>
<p><strong>further clarification</strong></p>
<ul>
<li>Yes, "press nut" was the word I'm searching for. </li>
<li>Yes the nuts I want to use need holes drilled/punched/lasered into the sheet before insertion. </li>
<li>I probably need different nuts of different sizes and threading, so I'm here for an abstract solution</li>
<li>I know the hole wizard. I know how to place a part into an assembly and to align it.</li>
</ul>
<p>What I want is this behaviour:
Upon manually placing a press nut into an assembly and aligning it in its recommended position on a metal sheet, the necessary hole shall be generated automatically in the metal sheet. </p>
|
|modeling|solidworks|
|
<p>I can't tell what you want. </p>
<blockquote>
<p>A combination of a part and forming tool</p>
</blockquote>
<p>What? </p>
<blockquote>
<p>I want to drag this part into a closed sheet metal area and the holes for the nut should be created automagically</p>
</blockquote>
<p>So, you want the hole wizard? Go to the hole wizard, pick the size hole you want, then go to the "Positions" tab, click a face and you're able to drop holes of the spec you chose wherever you want. </p>
<p>You linked the entire PEM nut catalog for press nuts (as <a href="https://engineering.stackexchange.com/a/12551/1633">S Barry correctly called them</a>), so I don't know which one you're talking about specifically. </p>
<p>That said, as far as I know, all of those nuts have you drill one hole in the sheet. Specification varies by nut, but if you <a href="http://catalog.pemnet.com/viewitems/floating-self-clinching-fasteners-a4-as-ac/floating-self-clinching-fasteners-a4-as-ac-unified" rel="nofollow noreferrer">look at any of the pages</a> on the site you linked, for a specific nut, you can see the hole size and tolerance you need to specify to the sheet metal fabricator. It's that big column labeled, "Hole Size in Sheet." </p>
<p>You can specify a custom hole size and tolerance when you're in the hole wizard. If you want to re-use that hole wizard setting repeatedly without having to type the values in every time, then you could <a href="http://blogs.solidworks.com/solidworksblog/2013/01/how-to-add-a-custom-hole-size-to-the-hole-wizard-database-in-solidworks.html" rel="nofollow noreferrer">add the custom hole settings to the wizard</a> to save those settings. </p>
<p>If this isn't the functionality you're looking for, or if you have a specific part you're referring to that couldn't use this method then please link to the specific part.</p>
|
12521
|
How to model a sheet metal nut in SolidWorks?
|
2016-11-27T19:46:33.890
|
<p>I'm looking to create something like this:
<a href="https://i.stack.imgur.com/XSZiL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XSZiL.png" alt="enter image description here"></a></p>
<p>But the problem with solid copper wires is it tends to spring back into it's original shape and this construction will flatten.</p>
<p>I've tried creating a wooden cone and wrap the copper around it; however, after leaving it for 24 hrs, it still flattens when I take it off the wooden cone.</p>
<p>Any help is appreciated, thanks!</p>
|
|metallurgy|metals|
|
<p>When you bend copper, it becomes much more "springy" than its original state. The wire will almost certainly have been bent at some stage before you use it, so the first thing you need to do it get it back to its original "soft" state. </p>
<p>You do that by <em>annealing</em> the wire. Heat it with a gas torch till it is red hot, and allow it to cool in air. </p>
<p>After it is annealed, make sure you only bend it <strong><em>once</em></strong>, into its final shape. You can repeat the annealing and bending as many times as you like, but you can only make one bend that will "hold its shape" accurately each time you anneal the wire.</p>
<p>Leaving the wire wrapped around the cone for 24 hours (or even for a year) won't change anything.</p>
<p>Here are a some of videos of annealing copper and bending a spiral which might help. You could try bending a spiral, annealing the wire again, and then "expanding" the spiral to make the cone shape.</p>
<p><a href="https://www.youtube.com/watch?v=3jSz11lz8MA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=3jSz11lz8MA</a></p>
<p><a href="https://www.youtube.com/watch?v=n_p4bNZv1lM" rel="nofollow noreferrer">https://www.youtube.com/watch?v=n_p4bNZv1lM</a></p>
<p><a href="https://www.youtube.com/watch?v=zvg1CdbBQvI" rel="nofollow noreferrer">https://www.youtube.com/watch?v=zvg1CdbBQvI</a></p>
<p>You might also get some ideas from websites and videos on bonsai - bent copper wire is used to "train" the tree branches into the desired shapes.</p>
|
12531
|
How to create intersected conical copper coils?
|
2016-11-28T01:06:09.933
|
<p>In the notes , the author stated that when the force is applied through the centorid of cross section , the channel will bend and twist.
but , on the second page , the author stated that the shear center lies on an axis of symmetry of member's cross sectional area...
So, i am confused whether the member will twist or not when the force P is applied thru the centroid or shear center ....</p>
<p>Why the shear center is located at O , which is located outside of the C channel ?
<a href="https://i.stack.imgur.com/MEH1W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MEH1W.png" alt="iamge2"></a></p>
<p><a href="https://i.stack.imgur.com/HSc0B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HSc0B.png" alt="image"></a>
Is there anything wrong with the notes ?</p>
|
|structural-engineering|
|
<p>Nothing wrong with the book or notes. This is just a little unintuitive at first until you take some time to internalize it.</p>
<p>The shear center lies on an axis of symmetry. If two axis of symmetry exist, it will lie on both of them at the centroid. Otherwise, it is only constrained to lie on one of them. The location of the shear center on the unsymmetric axis is determined by the equation for e. </p>
<p>If the centroid is located at the shear center, then it will not twist. However, if they are in different locations, twisting will happen at the centroid if a force is applied there.</p>
<p><a href="https://i.stack.imgur.com/dVYiS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dVYiS.png" alt="enter image description here"></a> </p>
<p>According to McGraw Hill Companies inc, </p>
<p>Shear Center: Of any cross section of a beam, that point in the plane of the cross section through which a transverse load must be applied in order that there will be only bending of the section and no twisting.</p>
<p>This picture shows how this works:
<a href="https://i.stack.imgur.com/hGi1g.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hGi1g.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/Y0272.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y0272.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/IJ1SK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IJ1SK.jpg" alt="enter image description here"></a></p>
<p>If the downwards force is applied along the screw at any point other than the shear center, twisting will occur. </p>
|
12535
|
twisting at shear center / centroid
|
2016-11-28T07:22:38.040
|
<p>I am trying to solve the following beam deflection problem:</p>
<blockquote>
<p>The cantilever beam is subjected to the point load at C.</p>
<ol>
<li>Generate the equation for the elastic curve by using the double integration method.</li>
<li>Find the maximum deflection and slope if L = 3 m and P = 10 kN acted at 2 m from A.</li>
</ol>
<p><a href="https://i.stack.imgur.com/K6SdQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K6SdQ.png" alt="A cantilever beam is fixed on the left (point A). Force P is shown acting in the downward direction at the rightmost end of the beam (point C). The length of the beam is L."></a></p>
<p><strong>Ans:</strong> $y_{max} = -46.67/{EI}$</p>
</blockquote>
<p>For part 1, my answer is $EI\dfrac{d^2y}{dx^2} = - Px$.</p>
<p>For part 2, my answer is:
$$\begin{gather}
EI\dfrac{dy}{dx} = -\dfrac{Px^2}{2} +c_1 \\
EIy = -\dfrac{Px^3}{6} +c_1x +c2
\end{gather}$$</p>
<ul>
<li>at $x= 0$, $y = 0$, so $c_2 = 0$</li>
<li>at $x = 0$ , $\dfrac{dy}{dx} = 0$, so $c_1 = 0$</li>
</ul>
<p>So, $EIy = -\dfrac{Px^3}{6}$.</p>
<p>$EIy$ max occurs at $L=3$, so $EIy_{max} = -\dfrac{10\cdot3^3}{6} = -45$, but the answer is $EIy_{max} =-46.67$. What have I done wrong?</p>
<p>For the slope at x = 2, my answer is $EI\dfrac{dy}{dx} = -\dfrac{Px^2}{2} = -\dfrac{10\cdot2^2}{2} = -\dfrac{40}{EI}$.</p>
|
|structural-engineering|beam|deflection|
|
<p>Both answers are incorrect.</p>
<p>Your answer to (a) would imply that the bending moment is equal to zero at $x=0$ and increases linearly until $x=L$. This is only true if you put $x=0$ at the free end of the beam and $x=L$ at the fixed support, in which case you'd have to change your constraints. What you actually need is:</p>
<p>$$EI\dfrac{d^2y}{dx^2} = -PL + Px$$</p>
<p>For (b), you didn't notice that the behavior is different in such a beam. You can't simply use the equation above and be done with it. After all, this equation describes a beam under load, but the span from $x \in [2, 3]$ is not under load and therefore its beam equation is different (it follows a linear path).</p>
<p>The easy way to solve it is to start by finding the slope and deflection at $x=2$. For this, we can use the equation above:</p>
<p>$$\begin{gather}
EI\dfrac{dy}{dx} = -PLx + \dfrac{Px^2}{2} + c_1 \\
\dfrac{dy}{dx}(x=0) = 0 = c_1 \\
EIy = -\dfrac{PLx}{2} + \dfrac{Px^3}{6} + c_2 \\
y(x=0) = 0 = c_2 \\
\therefore EI\dfrac{dy}{dx} = -PLx + \dfrac{Px^2}{2} \\
\therefore EIy = -\dfrac{PLx^2}{2} + \dfrac{Px^3}{6}
\end{gather}$$</p>
<p>From which we gather that</p>
<p>$$\begin{align}
\dfrac{dy}{dx}(2) &= -\dfrac{20}{EI} \\
y(2) &= -\dfrac{26.667}{EI}
\end{align}$$</p>
<p>From $x = 2$ to $x = 3$, the deflection is linear, meaning that the slope is constant, so the result above for $\dfrac{dy}{dx}$ is the same at $x=3$. Knowing this, we can find the deflection at $x=3$ by finding the linear component and then adding it to $y(2)$ (where $\ell$ is the linear span):</p>
<p>$$\begin{align}
\delta &= \dfrac{dy}{dx}(2)\cdot\ell = -\dfrac{26.667}{EI}\cdot1\\
\therefore y(3) &= y(2) + \delta = -\dfrac{46.667}{EI}
\end{align}$$</p>
|
12538
|
How to calculate deflection of a cantilever beam subject to point loading
|
2016-11-28T23:29:59.473
|
<p><a href="https://i.stack.imgur.com/uRKBT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uRKBT.png" alt="image1"></a></p>
<p><a href="https://i.stack.imgur.com/LIDbv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LIDbv.png" alt="im2"></a></p>
<p>in the first photo , it's stated that the dy/dx at center = 0 , however , in the second photo (website) , it's stated that the dy/dx at the boundary = 0 ,which is correct ? I'm confused now
<a href="http://www.geom.uiuc.edu/education/calc-init/static-beam/support.html" rel="nofollow noreferrer">http://www.geom.uiuc.edu/education/calc-init/static-beam/support.html</a></p>
|
|structural-engineering|
|
<p>If $\theta$ is supposed to be the same quantity as $dy/dx$, the picture is both drawn and labelled wrongly.</p>
<p>The equations correctly state that at $x=2$, $[dy/dx]_\text{AC} = [dy/dx]_\text{BC}$, but that does not mean $[dy/dx]_\text{AC} = [dy/dx]_\text{BC} = 0$!</p>
<p>On the other hand, $\theta$ might be some other quantity that isn't defined in your post - in which case we can't guess what it means.</p>
|
12553
|
slope of deflection of beam at center
|
2016-11-29T04:51:35.987
|
<p>In Euler's bending theory, it's stated that the critical stress of a beam is always larger than the yield stress of the beam. I don't really undertstand the difference between them. Critical stress is defined as the stress that the maximum stress applied before the beam starts to buckle.</p>
<p>Does the yield stress here mean the stress which the beam start to deform plastically? Why shouldn't the yield stress be smaller than the critical stress? </p>
<p>When the object is subjected to stress, the beam will deform plastically before it starts to bend and break. So I think that yield stress is smaller than the critical stress.</p>
|
|structural-engineering|
|
<p>Buckling and yielding are completely different concepts.</p>
<p><em>Buckling</em> occurs when an elastic structure does not move back to its initial position as long as the load is applied. Buckling is a purely elastic phenomenon. When elastic material is subjected to a small load, the deformation is usually small as well. This is different with buckling. Buckling requires a significant load in one direction, but then a very, very small load in a perpendicular direction can lead to destructive deformation (depending on the amount of deformation allowed by the system). Buckling is a stability problem, and the sample geometry is essential.</p>
<p><em>Yielding</em> occurs when the behavior of the material itself changes (due to the high load). When a material yields, the relative position of the atoms change.</p>
<p><strong>Edit:</strong> To realise the 'very, very small load' in perpendicular direction it is usually sufficient to apply the large load not perfectly in line with the stability axis. Since buckling is a stability problem, the large load leads to an instable system in the first place. In order to actually deform, some initial deformation in perpendicular direction is required which is then amplified. </p>
<p>I also slightly changed the first sentence of the buckling paragraph to account for Wasabi's comment.</p>
|
12558
|
yield stress vs critical load stress
|
2016-11-29T14:28:20.993
|
<p>How can water be removed from a submarine that is submerged?</p>
<p>If a submarine is opened, while it is submerged, water would get in because of the difference between the pressures. </p>
<p>For example, I have seen that there is a diving room in submarines, these rooms are in direct contact with the sea but the level of the water remains steady.</p>
<p>Can anyone explain for me this phenomenon?</p>
<p>If just the water level was steady, why don't we threw the water directly to the sea?</p>
<p><a href="https://i.stack.imgur.com/MEWCe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MEWCe.jpg" alt="explanation"></a></p>
|
|fluid-mechanics|pressure|
|
<p>The picture that you have drawn is not of submarine but a diving bell (check up wikipedia). In a diving bell the pressure inside the chamber balances the hydrostatic pressure of water and hence the water level inside the diving bell does not rise. You have see this if you push a glass tumbler into a tub. </p>
<p>In a true submarine, the water in the balast tanks is vented by pushing in compressed air from reservoirs. However this is not the only way submarines can control their depth.</p>
|
12563
|
How to get remove water from a submarine that is submerged
|
2016-11-29T17:43:59.080
|
<p>Consider $$\hat{G}(s) = \frac{1}{s^2+s}$$
than the Nyquist plot is
<a href="https://i.stack.imgur.com/gD1aj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gD1aj.jpg" alt="nyquist plot"></a></p>
<p>and the Bode plot is
<a href="https://i.stack.imgur.com/BYylp.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BYylp.jpg" alt="Bode Plot"></a></p>
<p>In both plots it seems that the closed-loop system is stable even when the eigenvalues are {0, -1}. For a higher order closed-loop system one can see that a system like </p>
<p>$$\hat{G}(s) = \frac{1}{s^3+s^2+s}$$</p>
<p>is at its stability limit since ne Nquist plots hits the real axis at -1 and in the Bode plot there is no phase reserve when magnitude hits 0. Why dose the Nyquist and the Bode plot fail for a closed-loop system of 2nd order?</p>
|
|control-engineering|control-theory|
|
<p>I think you are mixing up the closed- and open-loop systems.</p>
<p>The closed-loop system is $$\frac{\frac{1}{s^2+s}}{1+\frac{1}{s^2+s}}=\frac{1}{s^2+s+1}$$ </p>
<p>This has poles at $-0.5\pm 0.866025 i$ which is stable. </p>
|
12569
|
Why can’t one see for 2nd order system that it is at its stability limit neither in the Nyquist plot nor Bode plot?
|
2016-12-01T05:06:32.543
|
<p>I am an engineering student. I am working on a project where we need to use something called a vertical pipeline crawler. It is just a device that travels inside pipelines for inspection.</p>
<p>I found a crawler online, and I wanted to calculate the amount of power it uses. In specifications (there is more specs than below):
Speeds up to 10 m/min
Mass: 10 kg
Maximum pull: 27 kg
Power: 600 W 115 / 230 VAC</p>
<p>Initially what I did was to multiply the weight it can carry by the speed:
m<em>g</em>v = 27*9.81*(10/60) = 44.145 Watts
But it also says 600 W for power in specs. So how much power does this device need? Is the input 600 and output 44.145, isn't that very inefficient for an electrical device? Is my thought process wrong, what should use?</p>
<p>(I didn't want to share the link for the product thinking it might be against the website rules, however its pretty easy to find online)</p>
|
|mechanical-engineering|motors|power|
|
<p>Your question "power does this device need?" is unclear. If you are concerned about how much electrical power it requires, thats 600W and the current it requires will be given by dividing thus with the line voltage. The diameter of the electrical power cable will be determined by this.</p>
<p>When you multiply speed with the downward force you are neglecting how much energy the device has to expend in operating the gripping mechanism. In addition there is the electrical power requirement for the inspection head. You need to consider all of these.</p>
|
12592
|
Calculation of power for a pipeline crawler
|
2016-12-01T15:01:45.447
|
<p>I am a researcher on robotics and AI, and wonder if traditional manufacturing machines, such as lathe, CNC, or grind use any kind of machine learning technique or algorithm. This algos are widely used in mobile robots.</p>
<p>I.e.: in a plant where there are some lathes, do they interact somehow between them? Do they adjust on the fly their parameters when machining a new piece? Or do lathes try to adjust to the parameters to the lathe that is performing the best? Kind of collaborative learning?</p>
<p>I did not find this kind of examples in industry yet, but I wonder if industry 4.0 is about implementing this kind of techniques.</p>
|
|control-engineering|manufacturing-engineering|robotics|
|
<p>In general terms what you really want from CNC machines is to repeat the same actions exactly the same every time, while you may need to monitor and account for things like tool wear, temperature changes, small variations in material properties etc this is still in the realms of fairly straightforward feedback control. </p>
<p>Ultimately lathes aren't the best example as they really only only really have 3 degrees of freedom, cutting speed, tool feed and tool travel. A better example would be 5 or 6 axis milling which is inherently a lot more complex. Here there may well be scope for AI applications, particularly in designing and developing tool path design. This is the process of specifying the sequence of cutting moves to machine a part as quickly and efficiently as possible to a specific quality standard and may require numerous tool changes as well as avoiding collisions, nesting parts, deciding on which order to perform operations, optimising part/billet setup and orientation. </p>
<p>In particular there is a lot to be gained in the paths used for roughing cuts and finding the optimum tool path to move material quickly and effectively for example by taking a smoothed path around sharp corners rather than just using a fixed offset from the final geometry. </p>
<p>Automated tool path design has been around for a while but it is certainly conceivable that AI could have a role to play and it is certainly an important field. </p>
|
12604
|
Artificial Intelligence in manufacturing machines
|
2016-12-01T20:26:14.687
|
<p>What is the difference between impulse, step and sum of sinusoids signals in terms of their spectra and their degrees of excitation ?</p>
|
|control-engineering|control-theory|systems-engineering|
|
<p>An impulse has a flat/constant power density for all frequencies. The impulse response of a system can also be used to characterize a system, namely convolution of the input with impulse response, will yield the system response (however usually you would do this in the frequency domain, since then convolution becomes simple multiplication). The downsides of using an impulse as a way to excite a system is that the amount energy you can input into the system will be limited, since a too big impulse might damage a physical system. This might mean that you need a sensor with a higher accuracy, to measure lightly damped modes with (very) small gains, otherwise this might go unnoticed due to the resolution of the sensor.</p>
<p>A step has a spectrum with a slope of minus one. So low frequency dynamics will get a higher gain than high frequency dynamics. And again a step can only supply a limited amount of energy into the system.</p>
<p>If you actually mean just a sum of a couple of sinusoidal signals, then you will mainly see the system response of the system at those frequencies, but also some leakage to other frequencies, which could excite eigenmodes of the system. After these eigenmodes have died out (assuming the system is stable and sufficiently damped) then you should only see the response of the system at those frequencies. This can be useful if you are only interested in a specific frequency range or to see if the system can be modeled as a linear time independent system, because nonlinear or time variant systems will most likely also contain different frequencies in the response after the transient has died out.</p>
<p>Another good options for signals are sine sweep, by applying a single "sinusoidal" signal whose frequency changes (relatively slowly) over time, and (white) noise. The sine sweep is similar to the sum of sinusoidal signals, but it might be harder to be used to detect non-LTI system behavior. White noise, similar to the impulse, also has a flat spectral power density, but has the advantage that it continuously adds energy to the system and keeps exciting all modes. But it would be even harder to detect non-LTI system behavior.</p>
|
12608
|
Control System Question - spectra and their degrees of excitation
|
2016-12-03T14:41:24.447
|
<p>Hello I have a problem determining the value of K in this block diagram, to calculate this value, I applied this method, but the value at the end of K I is different from 216, you could explain to me where I'm wrong</p>
<p><a href="https://i.stack.imgur.com/rRkmf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rRkmf.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/edZs5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/edZs5.png" alt="enter image description here" /></a></p>
|
|mechanical-engineering|control-engineering|
|
<p>So, for this problem, you first need to find the transfer function from U to E. An easy way to do this to use the forward gain over one plus the loop gain. This simply mean $\frac{E(s)}{U(s)} = \frac{forward~gain}{1+loop~gain}$.</p>
<p>The forward gain in this problem is 1 (U to E has no gains between it) while the loop gain is $\frac{K}{(s+2)(s+3)}*\frac{1}{s+4}$ (all gains around loop from E back to U).</p>
<p>The steady state error is $e_{ss} = \lim_{s\to0} s\frac{E(s)}{U(s)} U(s)$. This simplifies to what you had above. If you substitute everything is, you have $0.1 = \lim_{s\to0}\frac{1}{1+\frac{K}{(s+2)(s+3)(s+4)}}$. </p>
<p>Multiply both top and bottom by $(s+2)(s+3)(s+4)$ and set s equal to 0. Multiply it all out and you should get 216.</p>
|
12630
|
Automation: steady state error calculation block diagram
|
2016-12-03T17:55:30.227
|
<p>I hope I'm asking this in the right place.
I have the following problem:</p>
<p><a href="https://i.stack.imgur.com/Jq2lf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jq2lf.jpg" alt="enter image description here"></a></p>
<p>and am to calculate A,B & C so that the system is in static equilibrium.
I calculated A without too many problems.</p>
<p>My problem is with B & C</p>
<p><strong>What I have done so far:</strong></p>
<p>I realised that my placing my moment at the point ABF I can eliminate B as an unknown and solve for C. </p>
<p>I calculated the resultant for the diagonal part ($R_1=q_0\cdot l$) and for the vertical part ($R_2=\frac{q_0\cdot 2l}{2}=q_0\cdot l$).</p>
<p>I calculated the distances of $R_1$ and $R_2$ from my moment $M^{ABF}$ using the Pythagorean theory and set them both negative as they are turning my system in a mathematically negative direction.</p>
<p>The force C I added to my moment as $C\cdot 2l$.</p>
<p>My moment calculation is at this point:</p>
<p>$$\Sigma M^{ABF}=2l\cdot C-(\sqrt{(2l)^2+(2l)^2}+\frac{l}{2})\cdot R_1-\sqrt{(2l)^2+(\frac{2}{3}2l})^2\cdot R_2$$</p>
<p><strong>My problem:</strong>
I don't know what to do with the moment $M$ which is at a distance of $l$ from $M^{ABF}$. I have tried adding as it is, adding it and multiplying by $l$, but both times the answer comes out wrong (I have a numerical answer).</p>
<p>I'm obviously doing something wrong but I don't know what.</p>
<p>In case it helps to have values:<br>
$F=4q_0l$, $M=2q_0l^2$, $\alpha=45°$<br>
$q_0=3kN/m$, $l=2m$</p>
|
|statics|
|
<p>Bending moments due to forces and distances are found by the product of the force and the perpendicular distance of its projection to the point of study.</p>
<p>You have the following structure (ignore the dimensions, they are just placeholders):</p>
<p><a href="https://i.stack.imgur.com/bKgP0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bKgP0.png" alt="enter image description here"></a></p>
<p>Transforming the distributed loads into equivalent concentrated loads, you get:</p>
<p><a href="https://i.stack.imgur.com/jBQ8E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jBQ8E.png" alt="enter image description here"></a></p>
<p>That figure already shows you the dimensions you need to consider for each force:</p>
<ul>
<li>The linear load along the <span class="math-container">$2\ell$</span> bar is horizontal. Therefore, the perpendicular distance to use is the vertical distance.</li>
<li>The uniform load along the <span class="math-container">$\ell$</span> bar is at a 135 degree angle with the horizontal. Therefore, the distance must be at a 45 degree angle.</li>
</ul>
<p>The concentrated bending moment <span class="math-container">$M$</span> in the middle of the horizontal <span class="math-container">$2\ell$</span> bar is simply added to your sum. Such loads are always just added to the equilibrium equation.</p>
<p>So, we can get:</p>
<p><span class="math-container">$$\begin{align}
M_{ABF} = 0 &= M + 2C\ell - \dfrac{2q_0\ell}{2}\cdot\dfrac{2}{3}2\ell - q_0\ell\left(2\ell\sqrt2 + \dfrac{\ell}{2}\right) \\
&= M + 2C\ell - q_0\ell^2\left(\dfrac{4}{3} + \left(2\sqrt2 + \dfrac{1}{2}\right)\right) \\
\therefore C &= \dfrac{q_0\ell^2\left(\dfrac{4}{3} + \left(2\sqrt2 + \dfrac{1}{2}\right)\right) - M}{2\ell} \\
\end{align}$$</span></p>
<p>Knowing this, it is then trivial to find the value of <span class="math-container">$B$</span> via the equilibrium of vertical forces <span class="math-container">$\left(\sum F_y = 0\right)$</span> equation. This is left as an exercise for the reader.</p>
<p><sub>All diagrams were done with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis tool.</sub></p>
|
12634
|
Static equilibrium and moment
|
2016-12-03T18:38:49.653
|
<p>It may be a simple or obvious question, but I'm confused what's changed between those. </p>
<p><a href="https://i.stack.imgur.com/fSDbM.png" rel="noreferrer"><img src="https://i.stack.imgur.com/fSDbM.png" alt="enter image description here"></a></p>
<p>That would be an example of a non structural rivet, does the bottom half of the mandrel steam inside the hollow rivet fall out, unlike structural where it stays, thus strengthening the rivet due to it not being hollow?</p>
<p>Bit confused between both.</p>
|
|rivets|
|
<p>Pop Rivets are a fastener composed of two portions: the mandrel and the hat. To install a pop rivet a hole is drilled into the materials, and then the hat is placed into the hole. A rivet installation too is then slid onto the exposed mandrel end. As it begins pulling the mandrel, it deforms the back of the hat creating a strong and quick hold. Once a certain pressure is reached, the mandrel snaps leaving the hat deformed and in place.</p>
<p>Basically, and using the terminology from your diagram, the <strong>pulling stem</strong> is pulled until exceeding its capacity. During this pulling period the <strong>rivet shell</strong> deforms and creates a <strong>header</strong> on the rear of the installation. Once the pulling stem is pulled too tightly it will snap resulting in the <strong>stem break point</strong> being just below the surface of the rivet shell. This creates a firm secure hold.</p>
<p>Pop rivets are most commonly used for their quick installation process and because you only need access to one side of the installation materials earning them the name blind rivets.</p>
<p>Structural rivets are a bit different as there is a built in locking mechanism that ensure the mandrel portion that stays inside the rivet does not fall out but actually creates a secondary wall. This increases both the rivets strength and hold over time.</p>
|
12636
|
What's the difference between structural pop rivets, and non structural?
|
2016-12-04T13:06:28.790
|
<p>I thought of an injector nozzle for a diesel engine with an alternate combustion chamber format (somewhat like a <a href="https://en.wikipedia.org/wiki/Heron_cylinder_head" rel="nofollow noreferrer">Heron combustion chamber</a>), and would now like to get an estimate of the size of the opening for it.</p>
<p>The injector nozzle is basically a tube with a pintle inside it; the pintle has a flange at its end, outside of the tube and in the combustion chamber, which extends past the radius of the tube. When the injector is spraying, the fuel flows between the tube wall and pintle and is deflected outwards by the flange into the combustion chamber, forming a planar 'disk' of fuel between the piston and the top of the cylinder. The problem is to calculate an estimate of the opening $l$ between the flange and the butt end of the tube from a simplified description of the intended spray front.</p>
<p><a href="https://i.stack.imgur.com/Qhk06.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qhk06.jpg" alt="Stylized illustration of the nozzle"></a></p>
<p>I'll be using cylindrical coordinates $r,\theta,z$.</p>
<p>Design parameters:</p>
<ul>
<li>$Q_0$, the mass flow rate in the tube</li>
<li>$r_0$, the flange radius (and thus of the nozzle)</li>
<li>$r_c$, the radius of the combustion chamber</li>
<li>$t_c$, the time allotted for fuel injection at max RPM.</li>
</ul>
<p>Problem domain:</p>
<ul>
<li>$S(r)=2\pi(r-r_0) l$ (spray front)</li>
<li>$V(r)=\pi(r^2-r_0^2) l$ (volume occupied)</li>
</ul>
<p>Assumptions:</p>
<ul>
<li>The fuel occupies the whole opening of the nozzle, doesn't diverge vertically and distributes itself uniformly ($\frac{\partial\rho}{\partial r} = \frac{\partial\rho}{\partial\theta} =\frac{\partial\rho}{\partial z} = 0$)</li>
<li>$Q_0$ is the flow not just in the nozzle, but also at the spray front ($Q_0=\iint_S \mathbf j\cdot d\mathbf S$)</li>
</ul>
<p>So,</p>
<p>$$m(t_c)=\pi\rho(r_c^2-r_0^2) l=Q_0 t_c\\
\implies l=\frac{Q_0 t_c}{\pi\rho(r_c^2-r_0^2)}$$</p>
<p>I get $l$ = 1.34×10<sup>-5</sup> cm = 134 nm for an application in a high-speed two-stroke ($r_c$ = 2.2cm, $r_0$ = 1mm, $t_c$ = 2×10<sup>-4</sup>s (more or less the time taken to traverse 10°, my estimate of the injection duration, at 8000 rpm), $Q_0$ = .8 g/s), which <strong>is</strong> absurdly tiny.</p>
<p>Even though the order of magnitude of the result comes mainly from the injection duration, I don't trust my results and think that it comes from the assumptions I made. On top of that, I don't know how to take into account the end of injection prior to combustion, so I assumed the injector stops injecting at combustion.</p>
<p>Is there anything wrong with the analysis or calculation that I did?</p>
|
|geometry|diesel|
|
<p>Revisiting the problem anew, it turns out that the units were interacting in a way I wasn't taking into account.</p>
<p>When I asked this question, I had input $r_c$ and $r_0$ in cm. However, $\rho$ was in <em>kg/m<sup>3</sup></em>, and $Q_0$ was in <em>g/s</em>; additionally, my $Q_0$ was too large (an <a href="https://en.wikipedia.org/wiki/Brake_specific_fuel_consumption#Examples_of_values_of_BSFC_for_shaft_engines" rel="nofollow noreferrer">Audi 3.3L V8 TDI</a> achieves 0.014 g/s per injector).</p>
<p>Surely enough, putting $r_c$ and $r_0$ into meters, then putting $\rho$ in g/m<sup>3</sup> and finally correcting $Q_0$ gave me $l$ = 2.9342×10<sup>-6</sup>m ≈ 2.93 μm, which is more reasonable given that $2\pi r_0 l$ = 9.2181×10<sup>-8</sup> m<sup>2</sup> ≈ 9.22 μm<sup>2</sup>, comparable to the injection area per injector for a <a href="https://fr.wikipedia.org/wiki/HDi#Syst.C3.A8me_d.27injection" rel="nofollow noreferrer">PSA HDi DW10D</a> with 8×110-micrometer nozzles of 7.6 μm<sup>2</sup>.</p>
<p>If there's a lesson to be learned here, it's to double check not just your dimensions, but your units... especially the ones indirectly related to the result.</p>
|
12653
|
Diesel injector nozzle geometry
|
2016-12-04T13:43:35.990
|
<p>I'm having trouble with this problem:</p>
<p><a href="https://i.stack.imgur.com/OP64K.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OP64K.jpg" alt="problem"></a></p>
<p>G and a are given, and we are asked to calculate A, B and C. </p>
<p><strong>What I've Done:</strong><br>
The first thing I tried was a free body diagram cutting through A and C to create one body and C and B to create another. </p>
<p><a href="https://i.stack.imgur.com/vvHTr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vvHTr.jpg" alt="Body 1"></a></p>
<p>From here I calculated $A_y$ and $C_y$ as being equal to $-G\frac{\sqrt{2}}{4}$<br>
Then I moved onto the second body:</p>
<p><a href="https://i.stack.imgur.com/oRfgQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oRfgQ.jpg" alt="Body 2"></a></p>
<p>Here I ran into problems immediately. If I placed the moment at either B or C then it follows that the x component of the other must be zero, which it isn't (I have a numerical solution). So I've either made a mistake with my free body diagram or I'm doing something wrong with the second moment.</p>
<p>Here are my equalibrium calculations:<br>
Body 1:<br>
$$\Sigma M^{A}=C_y\cdot 2a+G\frac{\sqrt{2}}{2}=0 \\
\Sigma F_x=A_x-C_x+G\frac{\sqrt{2}}{2}=0 \\
\Sigma F_y=A_y+C_y+G\frac{\sqrt{2}}{2}=0$$ </p>
<p>Body 2:
$$\Sigma M^{B}=3a\cdot C_x=0 \\
\Sigma F_x=C_x+B_x=0 \\
\Sigma F_y=B_y-C_y=0$$ </p>
<p>My results obviously make no sense. What am I doing wrong? I have also tried to calculate for the whole structure but there were too many unknows which I couldn't get rid of.</p>
|
|statics|
|
<p>Let's start by transforming the load <span class="math-container">$G$</span> into the resultant forces applied to the structure. At point <span class="math-container">$R$</span>, we have a vertical component equal to <span class="math-container">$G$</span> and an inclined component <span class="math-container">$G$</span>. Since the inclined component is at 45°, we get a total force equal to <span class="math-container">$G\left(1 + \dfrac{1}{\sqrt2}\right)$</span> downwards and <span class="math-container">$\dfrac{G}{\sqrt2}$</span> to the left. Relatedly, at the end of the rope (between <span class="math-container">$A$</span> and <span class="math-container">$C$</span>), the concentrated force has upwards vertical and rightward horizontal components both equal to <span class="math-container">$\dfrac{G}{\sqrt2}$</span>.</p>
<p>The model is therefore equal to (for <span class="math-container">$G = 100$</span>):</p>
<p><a href="https://i.stack.imgur.com/QdReC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QdReC.png" alt="enter image description here"></a></p>
<p>As you already found, we can isolate <span class="math-container">$AC$</span>. It behaves like a simply supported beam with one horizontal constraint at <span class="math-container">$A$</span>. Therefore, the horizontal load is fully absorbed by <span class="math-container">$A_x$</span>. The vertical component is evenly split (due to being at the midspan of <span class="math-container">$AC$</span>) between <span class="math-container">$A$</span> and <span class="math-container">$C$</span>.</p>
<p>Now we have to deal with the forces at the top of the structure. The vertical force will be fully absorbed by <span class="math-container">$B$</span> (but is slightly offset by the vertical component at the end of the rope discussed above which is transmitted to <span class="math-container">$C$</span>). The horizontal force, however, we can't yet know. For that, we need to balance out the moment around <span class="math-container">$B$</span> considering these forces and the horizontal force applied on/by <span class="math-container">$AC$</span>.</p>
<p><span class="math-container">$$\begin{gather}\sum M_B = -G\left(1 + \dfrac{1}{\sqrt2}\right)\cdot3a + \dfrac{G}{\sqrt2}\cdot7a - A_x'\cdot3a = 0 \\
\therefore A_x' = G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3}
\end{gather}$$</span></p>
<p>where <span class="math-container">$A_x'$</span> is the horizontal reaction in <span class="math-container">$A$</span> due to those forces, to be added to the reaction found previously of <span class="math-container">$\dfrac{G}{\sqrt2}$</span>. So, <span class="math-container">$B_x = \dfrac{G}{\sqrt2} - A_x'$</span> (in this case <span class="math-container">$A_x'$</span> is negative, so this ends up being a sum).</p>
<p>So we end up with:
<span class="math-container">$$\begin{alignat}{3}
A_x &= -\dfrac{G}{\sqrt2} + G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(\dfrac{1}{3\sqrt2} - 1\right) &&\approx -0.764G \\
A_y &= -\dfrac{G}{2\sqrt2} &&&&\approx -0.354G \\
B_x &= \dfrac{G}{\sqrt2} - G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(1 - \dfrac{1}{3\sqrt2}\right) &&\approx 0.764G \\
B_y &= G\left(1 + \dfrac{1}{\sqrt2}\right) - \dfrac{G}{2\sqrt2} &&= G\left(1 + \dfrac{1}{2\sqrt2}\right) &&\approx 1.354G
\\ C_x &= G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(\dfrac{4}{3\sqrt2} - 1\right) = &&\approx -0.057G \\
C_y &= -\dfrac{G}{2\sqrt2} &&&&\approx -0.354G \\
\end{alignat}$$</span></p>
<p>And now, to check our work:</p>
<p><a href="https://i.stack.imgur.com/W1Tg0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W1Tg0.png" alt="enter image description here"></a></p>
<hr>
<p>Alternatively, using your method, what you forgot to do was to consider the bending moment due to the forces at the top of the structure in your second cut (<span class="math-container">$BC$</span>). That moment is equal to <span class="math-container">$G\left(1 + \dfrac{1}{\sqrt2}\right)3a - \dfrac{G}{\sqrt2}\cdot4a = Ga\left(3 - \dfrac{1}{\sqrt2}\right)$</span>, which can only be balanced by the force binary generated by <span class="math-container">$A_x$</span> and <span class="math-container">$B_x$</span>. Dividing the moment by the <span class="math-container">$3a$</span> lever arm between <span class="math-container">$A_x$</span> and <span class="math-container">$B_x$</span>, we get that each of those forces must be equal to <span class="math-container">$\pm\dfrac{Ga\left(3 - \dfrac{1}{\sqrt2}\right)}{3a} = \pm G\left(1 - \dfrac{1}{3\sqrt2}\right)$</span>, which is precisely what we got for <span class="math-container">$A_x$</span> and <span class="math-container">$B_x$</span> above. Knowing this, you can then solve for the other variables as well.</p>
<p><sub>All diagrams given by <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis tool.</sub></p>
|
12654
|
Free Body Diagram Problem
|
2016-12-04T21:24:31.650
|
<p>What is the difference between 250W band heater and a 200W one? Can they reach the same temperature?</p>
|
|heat-transfer|power-electronics|
|
<p>Heating elements generally have a maximum operating temperature above which they'll start to degrade and eventually break. If these heaters are from the same product line they are likely made of the same materials and will have the same maximum operating temp therefore.</p>
<p>Separately, you want to know how hot they can get in your application. The hotter they get, the faster they lose heat to the environment. Once they are losing heat as quickly as they generate it, their temperature will stop rising. From that perspective, the 250 W heater will reach a higher temperature if all other things are the same in your system. Alternatively, you may add insulation to the lower power heater to achieve the same final temp.</p>
<p>The last thing to consider, is that the higher power heater will heat things up more quickly.</p>
|
12663
|
What is the difference between 250W band heater and a 200W one?
|
2016-12-05T06:18:30.933
|
<p>I am looking for a method changing the coordinate system of RBE2 or RBE3 elements. </p>
<p>Below is the nastran description of RBE2 in Nastran user guide.</p>
<p><a href="https://i.stack.imgur.com/Z97bs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z97bs.png" alt="enter image description here"></a></p>
<p>There is no field for coordinate configuration.</p>
<p>Now, I am wondering if coordinate change is possible or not.</p>
<p>Could anyone explain how to change coordinate of RBE2?</p>
|
|mechanical-engineering|modeling|simulation|
|
<p>The RBE2 element uses the output coordinate systems that are defined for the grid points (field CD on the GRID card).</p>
<p>If you want to use two different coordinate systems at the same point in your model, you can define two grid points with different output coordinate systems, and join them with a zero-length RBE2 or RBAR element. </p>
<p>Note that if a grid point is used in more than one rigid element, it can only be a <em>dependent</em> grid point in <em>one</em> of those elements. But if the rigid element connects all 6 degrees of freedom at each grid, it makes no difference which grid you choose as the independent one, so this isn't a "real" restriction on what you can do - it's only a feature of the way NASTRAN works internally!</p>
|
12673
|
Nastran RBE2 coordinate change
|
2016-12-06T07:56:58.303
|
<p>A metallic cube of side 10cm, density 6.8gm/cc is floating in liquid mercury (density 13.6gm/cc), with 5cm height of cube exposed above mercury level. Water is filled over this to submerge the cube fully. What is the new height of the cube exposed above the mercury level?</p>
|
|fluid-mechanics|
|
<p>Question was really simple, <strong>We need to equate the total buoyancy force with the weight of the body</strong>. I was stuck with before and after situation so was not able to solve it. Answer is the block will rise by 0.4cm.</p>
<p>Total Buoyancy force for a body immersed in two fluid
= Density of water*g(S1*V1+S2*V2) </p>
<p>V1 & V2 being volume of block in respective fluid. S1 and S2 specific gravity</p>
<p>In this question 1-Mercury, 2-water</p>
<p>When water is filled over mercury, let us assume that the block will rise by a x units.</p>
<p>Buoyancy force= 1000*g(S1*(5-x)+S2(5+x))= Weight of block.</p>
|
12698
|
Body submerged in two liquids
|
2016-12-06T16:19:40.757
|
<p>I have a friend who is German and he says that wire mesh is much stronger than steel bars (rebar) and is mandated in Germany for reinforcing concrete slabs.</p>
<p>The question is, is that true, and why?</p>
<p>Here's a typical concrete reinforcement here (in Indonesia)</p>
<p><a href="https://i.stack.imgur.com/mjRdh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mjRdh.jpg" alt="rebar"></a></p>
<p>There are tied rebar beams, with rebar for floor reinforcement. Instead of the floor bars, we could instead install mesh between the beams. But what is the difference? Obviously we could buy 10.0 mm wire mesh @ 15 cm spacing, or we could install 10.0 mm bars at the same spacing (and let's say for the sake of argument that the same steel is used to make the mesh as we receive as bars). This would result in the same weight of steel per m<sup>3</sup>, However whereas the wire mesh is welded, the rebars are only tied with wire, which I guess has trivial strength.</p>
<p>[note, let us assume, for the sake of argument that the mesh and bars are properly placed]</p>
<p>Is there, in any case, an overwhelming advantage from the welds in the mesh, or can such advantage as there is be eliminated at relatively low cost, say by increasing the numbers of bars used by say 10%.</p>
|
|structural-engineering|concrete|reinforced-concrete|
|
<p>Wire mesh and rebar are roughly the same strength ; 55,000 psi yield ( both are available in various grades ). A heavy wire , 8 gage ( American Wire Gage) is 0.123" diameter . For an ordinary house , 0.5 " diameter rebar is common ( mine also has some 0.625" and 0.375" ) . Spacing is roughly the same , 6 in. between wire or rebar. So in a mesh structure there is 0.123" diameter steel and in a rebar structure there is 0.5" diameter steel in the same space. Without doing all the arithmetic; there is no way mesh is close to the strength of rebar. Mesh is used because it requires much less labor to install.</p>
|
12705
|
Relative strength of wire mesh vs rebar for reinforcing a concrete slab
|
2016-12-06T19:03:40.323
|
<p>Why is it necessary for the moment of the forces on a rigid body in static equilibrium to be zero about any point, even if the position of that point is not on the body itself.</p>
|
|statics|
|
<p>Consider a body to be in equilibrium under the action of forces $F_i$, $i=1,\ldots,n$ which act at locations $r_i$, $i=1,\ldots,n$.</p>
<p>Equilibrium of the body implies
$$\sum _{i=1}^n r_i \times F_i=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sum _{i=1}^n F_i=0$$</p>
<p>What about the moment about an arbitrary point p (that may or may not be on the body). </p>
<p><a href="https://i.stack.imgur.com/4fTKr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4fTKr.jpg" alt="enter image description here"></a></p>
<p>That can be computed as </p>
<p>$$\sum _{i=1}^n (r_i-p) \times F_i = \sum _{i=1}^n r_i \times F_i-p\times\sum _{i=1}^n F_i = 0-p\times 0 = 0$$</p>
<p>Hence the result. The moment is zero about any point, whether it is on the body or not.</p>
|
12707
|
Why is the moment of forces on a rigid body in static equilibrium must be zero about any point, even if the position of that point is not on the body
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.