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2016-03-11T11:22:28.093 | <p>I'm working on a flood defence scheme in which a section of RC flood wall becomes sheet piles. The reasons for this aren't important, but the solution isn't to simply make all walls one or the other. </p>
<p><strong>My issue is how to connect these two very different types of wall in a way that provides an expansion joint for the concrete but is also water-tight.</strong> This isn't an issue for connecting individual concrete panels as a waterstop is simply cast into the expansion joint, providing a seal. However, sheet piles don't have anything to 'cast' onto. </p>
<p>My idea so far is to design a sort of concrete (possibly mass, probably RC) 'pilaster' at the end of a section of sheet pile that will allow a water stop to be cast into. However, sheet piles deflect a fair amount which is making me wary of casting a rigid concrete section at the end of the sheet pile.</p>
<p>I'm at a bit of a loss here and struggling to find a precedent of a connection like this. SURELY it's been done before? </p>
<p>I've attached a quick hand sketch. I'm not too bothered about detailed calculations for this; I'm more interested in solving the buildability issues at the moment. </p>
<p>Dimensions are subject to minor alterations but roughly: </p>
<ul>
<li>The RC wall penetrates 1.5 m below ground level</li>
<li>Sheet pile penetrates 6 m below ground level</li>
<li>RC wall stem is ~350 mm thick</li>
<li>Sheet pile section is 450 mm thick (AU25 section)</li>
</ul>
<p><a href="https://i.stack.imgur.com/IQBlT.png" rel="nofollow noreferrer" title="Flood defence design sketch"><img src="https://i.stack.imgur.com/IQBlT.png" alt="" title="Flood defence design sketch"></a></p>
| |civil-engineering|geotechnical-engineering|reinforced-concrete|retaining-wall| | <p>We used a simple rubber between two concrete walls. I don't see any reason you can't use the same for a concrete steel wall connection. As these sorts of walls don't settle much you should be ok just coupling these two together without worrying about one settling and dragging the other. Just weld a flat sheet (possibly lined with rubber for water tightness) to the end of the steel one and lowered into the wet concrete. </p>
| 7879 | Connections between RC retaining walls and sheet pile retaining walls |
2016-03-11T13:53:04.200 | <p>A cascaded PID controller is depicted in following image (courtesy of <a href="http://www.controlglobal.com/assets/Media/0512/article_547_fig3.jpg" rel="nofollow noreferrer">controlglobal.com</a>):
<a href="https://i.stack.imgur.com/fCyjy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fCyjy.png" alt="enter image description here"></a></p>
<p>The primary PID tracks the setpoint $SP_1$ and delivers a setpoint $SP_2$ to the secondary PID.</p>
<p>I wondered if it could be a problem if the secondary PID does not have a nice behaviour w.r.t. tracking its setpoint, while the total result of the cascaded PID controller is OK. This would mean that the overall tracking of $SP_1$ is OK, which is the goal of the total controller, while some internal controllers are not tuned too well.</p>
| |control-engineering|control-theory|pid-control| | <p>In general, poor performance of PID2 will cause poor performance in PID1, in both disturbance rejection and setpoint tracking.</p>
<p>There are some special instances where poor performance of PID2 in the form of high frequency oscillation is attenuated by G1, so the performance of PID1 is not significantly affected. </p>
| 7885 | Is it important to have a good behaviour inside internal loops of a cascaded PID controller? |
2016-03-11T21:15:29.307 | <p>I have been trying to model a custom gear to fit on to another physical gear. I having been using <a href="http://www.rushgears.com/tech-tools/part-search/build-custom-gears" rel="nofollow">this 3d model gear generator</a>. What I know about my <a href="https://www.servocity.com/html/32_pitch_hitec_servo_gears.html#.VuMyM-IrLIU" rel="nofollow">physical gear</a> is that it has:</p>
<ul>
<li>(what I believe to be a diametral) Pitch of 32</li>
<li>Pitch Diameter of 1.5" (38.1mm)</li>
<li>Outer Diameter of 1.563" (39.7002mm)</li>
<li>20° Pressure Angle</li>
<li>48 Teeth</li>
</ul>
<p>I have access to a 3D printer and am trying to calculate a gear that will mesh with my physical gear given these specs. I printed a smaller first gear that happened to mesh perfectly with the physical gear. <strong>When I increased the number of teeth</strong> (thus increasing the PD and OD of the gear in the generator) <strong>the teeth and distance between the teeth appeared to increase as well</strong>, not meshing well with the physical gear at all.</p>
<hr>
<p>New gear specs:</p>
<ul>
<li>Pitch Diameter of 1.2" (30.48mm)</li>
<li>Outer Diameter (Auto calculated, but I think it should be) 1.2" + .063" (30.48mm + 1.6002mm)</li>
<li>Same Pressure Angle</li>
<li>*No blacklash</li>
</ul>
<hr>
<blockquote>
<p>Is there a way to calculate a gear that will mesh, or will I have to
empirically derive the gear tooth and pitch by printing out gear after
gear to see if it fits?</p>
</blockquote>
| |gears| | <p>Gears will only mesh together when they both share the same Diametral Pitch. There are many online descriptions for this term, that is what you want to learn about. Here is an example of one: <a href="https://www.bostongear.com/pdf/gearology/chapter02.pdf" rel="nofollow">https://www.bostongear.com/pdf/gearology/chapter02.pdf</a></p>
| 7896 | 3d model gear to fit another gear |
2016-03-11T21:35:03.963 | <p>I've heard that both of these coupling types introduce a degree of kinematic error in the driven shaft. What does it actually mean in practice? Is it that the driven shaft doesn't turn uniformly even when the input shaft does? But is this error arbitrary or the same amount of error is experienced for every revolution?</p>
<p>In fact, in what sort of application should one be used but not the other?</p>
<p>Any thought appreciated.</p>
| |mechanical-engineering| | <p>The short answer is that universal joints are tolerant of angular misalignment while Oldham couplings are intended for use where the shafts are misaligned but parallel.</p>
| 7898 | Oldham coupling vs Universal joint |
2016-03-11T23:18:31.990 | <p>I love ejection seats and the flying contraptions they usually reside in. In fact, I like them so much that I want to make my own, despite my profound lack of training and inexperience with mechanical engineering in general. After doing a little research into the materials for my ejection seat, I found a large collection of aluminum alloys to choose from. Some are fairly simple to eliminate like the 1000, 2000, and 3000 series. However, between the various grades in the 5000 and 6000 series, I have a lot more trouble.</p>
<p>I read that 6061 is the most common alloy for reasons of weldability, corrosion resistance and a few properties. Also, I intend to rivet the chair together because that's what I see in my detailed source pictures though I have a little experience with TIG welding so that may be an option too. </p>
<p><strong>In the benign environment of my office, will it really matter which alloy I pick? Can I just use common 6061 sheet or is there a better, cheaper alternative?</strong></p>
| |aluminum|welding|alloys|rivets| | <p>As an aside, if you want to bend and play with it. Here's a tip. Get a pot of soap, a gas blow torch and sit by a bath. Smear the part with soap. Heat with the blow torch until the soap turns black, then drop it in the cold bath. </p>
<p>Depending on your air temperature you've now got a few hours to bend it to your desired shape. once down leave it in the airing cupboard or somewhere else nice and toasty for a day or so. The alloy will then regain it's stiffness. </p>
<p>What you've done is melt the nickel and copper crystalline structures within the aluminium. Now it will fell, bend and act like pure simple aluminium. Once you give it time the copper and nickel will reform the internal structure and you're done.</p>
<p>Welcome to home heat treatment. it's exactly the same way I saw piper making skin panels in Vero Beach. Heat the cut to shape skin panels, drop them in a tank then bend them onto the back and clico rivet. Then they put the parts out in the Florida sunshine. Job done. </p>
| 7901 | Aluminum for an Ejection Seat Replica |
2016-03-13T19:42:53.723 | <p>The diagram is for three materials with different thermal conductivities. I think number 1 is the material with higher thermal conductivity because it transfers the heat faster so it gets to the steady temperature sooner than the other two. Am I right? If not or if my reasoning isn't correct, I appreciate your help.</p>
<p><a href="https://i.stack.imgur.com/nh1zL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nh1zL.jpg" alt="enter image description here"></a></p>
| |thermodynamics|heat-transfer|thermal-conduction| | <p>$\dot{Q} = k A \frac{\Delta T}{l}$</p>
<p>$l$ thickness </p>
<p>$A$ surface area</p>
<p>From this follows that besides other factors the rate of heatflow is proportional to the thermal conductivity $k$.</p>
<p>$\dot{Q} \propto k$ </p>
<p>Therefore a high thermal conductivity allows for a fast heat transfer, hence your interpretation that a fast change in temperature correlates with a high thermal conductivity is right.</p>
| 7915 | Relationship between thermal conductivity and reaching the steady temperature? |
2016-03-14T04:46:00.937 | <p>I'm running a simulation in a CAD on a rather simple assembly with several parts in which a motor effectively rotates a part. If the motor is set to constant 10RPM, the part rotates fine and the max. motor torque is around 500N-mm. However, if it's set to 5RPM, it gets stuck and the motor torque reaches 8000N-mm.</p>
<p>How come a higher RPM could avoid a lock-up? Isn't a higher torque less susceptible to rotational lock-ups, since it's the force to rotate an object?</p>
| |mechanical-engineering| | <p>Simulating a process is a big task which includes various preprocessing steps. We try to model the real world yet are not able to do it 100%. So sometimes it’s hard to find errors in your model. Please check for the following errors:</p>
<ul>
<li>Cad Modeling Error</li>
<li>Material Properties Error</li>
<li>Discretization Errors </li>
<li>Other small errors</li>
</ul>
<p>Check for the modeling error. There may be a problem in the geometry of parts and assembly of your motor. You should check the assembly mates and their constraints. Sometimes constraints are not given correctly which give wrong results while simulating the process.</p>
<p>Check your material properties which you have given to your motor parts. Improper or incomplete material properties will result in different behavior when you will simulate the process. For example, the limit of Poisson’s ratio in isotropic materials.</p>
<p>Your model may have Discretization errors. If you want to run the proper simulation with highly accurate results you can use the ALTAIR Hypermesh for meshing as there may be a possibility that your element size may be too large or too small. After meshing you can use the solver like ALTAIR RADIOSS or LS-DYNA. You can ANSYS workbench to simulate also. These software has better FEM solving capability and provides better meshing options than CAD software. You can use the mesh refinement method. Coarse meshing always gives the false results.
Apart from these, you should check for Hourglass effect, element type, contact options, time step, mesh type, damping controls, and erosion controls. Check for proper boundary conditions, if you forget to set boundary conditions you may get wrong results.</p>
<p>If you don’t find any error then try to use different CAE software for simulation. Sometimes CAD software are not able to properly simulate the process. If the error still persists you should consult with a CAD / CAE expert and show him your model. And keep in mind these simulations are approximations.</p>
| 7921 | Why does a motor lockup at low RPM in my simulation? |
2016-03-14T09:29:52.880 | <p>I'm trying to model a part that rotates on a shaft. I measured empirically that it has a static torque of around 500 N·mm and a dynamic torque of around 400 N·mm. I want to simulate it in CAD but my software only takes static/dynamic friction as inputs.</p>
<p>How can I convert the torque values to static/dynamic friction values for the part-shaft contact surface?</p>
| |mechanical-engineering|torque|friction| | <p>It is difficult to model with friction coefficients in this case, because the normal force is dependent on very small changes in dimensions and material strengths that your software may or may not support. Friction is by no means as easy as it is in the text books. The basic idea is find the force normal to the friction surface in question and calculate the coefficient with the <a href="https://simple.wikipedia.org/wiki/Coefficient_of_friction" rel="nofollow">friction equation</a>.</p>
<p>It may be more accurate and easier to just iterate to a solution. First you guess at a friction coefficient, insert it into your simulation, run your simulation, see if your result is higher or lower than amount you measured, then guess a new value and repeat until you feel that you are are close enough to your measured value.</p>
| 7924 | Determine coefficients of friction based on the torque required to rotate a part on a shaft |
2016-03-14T15:39:05.923 | <p>I have a circular puck of stainless steel that is 9.5 mm in diameter. In the center, there is already a 5 um pinhole. My objective is to basically cut the steel along the dotted line, approximately 0.5 mm from the pinhole. <strong>Can you advise on what techniques to can make this cut precisely?</strong></p>
<p><a href="https://i.stack.imgur.com/CMzB6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CMzB6.png" alt="Circular sheet of steel with 5 um pinhole"></a></p>
<p>The reason I want to do this is to be able to align two of these so that I can have 2 pinholes that are ~1 mm apart from one another, as in the following image.</p>
<p><a href="https://i.stack.imgur.com/WVdRk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WVdRk.png" alt="Two pinholes adjacent to one another"></a></p>
| |steel|cutting| | <p>CNC Milling should be precise enough (0.01 millimeter accuracy).</p>
| 7934 | How can I cut a thin (1mm) sheet of stainless steel with high precision? |
2016-03-15T04:32:55.640 | <p>I've been looking at solutions to compress gases and getting stunned by the cost and maintenance requirements of existing systems to fill cylinders (whether for SCBA or chemical storage) to 2000-4000psi: The cost of entry is over 4 figures, with commercial systems like those used in fire stations and SCUBA shops running into 5 figures. And, as far as I can tell, they are all <em>direct mechanically driven</em> multi-stage <em>pneumatic</em> pistons.</p>
<p>Meanwhile, sitting in my garage I have a manually-driven hydraulic bottle cylinder that can lift 20 tons, and typical hydraulic systems circulate hydraulic fluid at 5000-10000psi.</p>
<p>So I have in mind a very simple gas compressor: a single piston sealed on both ends. On one side it is driven by a single hydraulic pump line. On the other are two gas valves: An intake and an output. Operation is simple:</p>
<ol>
<li>Hydraulic pressure shut off.</li>
<li>Gas intake valve open.</li>
<li>Hydraulic flow reversed until piston has drawn in the maximum capacity of gas.</li>
<li>Intake valve closed.</li>
<li>Hydraulic pressure run until piston has reached maximum compression</li>
<li>Output valve opened (if not one-way and hence automatic), then closed once pressure equalized with output cylinder.</li>
</ol>
<p>If the piston lacks the compression ratio to reach the desired psi, then the charged cylinder is switched to the input line and another cylinder to the output.</p>
<p>Am I missing a reason this won't work? Or does it already exist in some form close to this?</p>
<p>If not, where and under what name might I find a long two-sided piston with pressure fittings on each end?</p>
| |hydraulics|compressors|compressed-gases| | <p>The elephant in the room is that while gasses are compressible, hydraulic oil is only negligibly so. </p>
<p>This difference has two major effects:</p>
<p>Firstly, the output volume flow is limited by the intake volumetric flow divided by the compression ratio. Let's say we have a hypothetical compressor that pulls in 10 L of air per second at STP, and raises the pressure to 300 bar. Once the output gasses have cooled down from temperatures best described as "red hot", the output volume will be 1/30th of a liter per second. All that work, and we only have 30 mL of compressed gas. </p>
<p>The second issue is that as you compress a gas it heats up. This is an effect that is inherent to the physics of compression. The heat (motion) already present in gasses doesn't go away when they are compressed. When our example 10 L (10,000 mL) of air is smashed into a much smaller volume of 33 mL, all the heat energy the air already contained remains, but in a much smaller volume. The air has been compressed to occupy a space that is 0.3% of its original volume - a size reduction of 99.7%. All of that heat has to be absorbed and dissipated by the compressor mechanism somehow before we will end up with cool high pressure gasses. </p>
<p>Keep in mind that air at 300 bar has an O2 partial pressure of around 63 Bar. High pressure O2 is extremely dangerous around oils or anything else flammable. For example, the oils like those used in and are essential to seal hydraulic systems. </p>
<p>Your energy source doesn't matter, whether hydraulic, electric, or steam. They are just how you apply the energy to the system, and make very little difference to the complexity of the compressor mechanism itself.</p>
| 7947 | Why no hydraulic piston gas compressors? |
2016-03-15T13:52:51.870 | <p>I am currently looking at a set of truss layouts produced by our truss manufacturing partner for a residential building project. I have reason to believe these layouts are generated using Alpine (Intelliview?) software. In any case, the design summary contains abbreviations which I cannot understand since I do not have a copy of the software. Please see below:</p>
<p><a href="https://i.stack.imgur.com/wCPtr.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/wCPtr.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/4HzmH.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/4HzmH.jpg" alt="enter image description here"></a></p>
<p>According to <a href="http://alpineitw.com/wp-content/uploads/2015/01/Read_A_Drawing.pdf" rel="noreferrer">this</a> document R, U and W stand for reaction due to design load, uplift due to wind load and bearing width, respectively. TCLL, TCDL, BCLL and BCDL stand for top chord live load, Top chord dead load, Bottom chord live load and bottom chord dead load. However I cannot for the life of me determine what Rw, Rh, RL and NCBCLL stand for. Any ideas?</p>
| |structural-engineering|structures|structural-analysis| | <p>NCBCLL is IBC Code for Non Concurrent Bottom Chord Live Load, used for uninhabitable/inhabitable attic space live loads @ 10/20 PSF through Alpine ITW</p>
| 7955 | Truss manufacturer drawings unclear |
2016-03-16T01:23:30.243 | <p>Al<sub>2</sub>O<sub>3</sub> is a hard, white material used for making firebricks and numerous molded ceramic parts.</p>
<p>Say I had a firebrick, took a core out of it using a thick-walled diamond hole saw and a drill press.</p>
<p>What lathe insert material would I then use to machine the outside of the Al<sub>2</sub>O<sub>3</sub> cylinder assuming I could take care of all the holes using diamond hole saws?</p>
<p>My budget for one cutting bit would be $500. But that would take me many hours of work to raise, so I prefer something less expensive, since I'm probably going to break it, hence <em>cost effective</em>.</p>
| |materials|machining|ceramics|lathe| | <p>In industrial applications probably the closest approximation is in dressing bonded abrasive wheels. These use abrasive particles including aluminum oxide but a different bonding medium which makes them a bit tougher than standard fire bricks. With this is mind a diamond tipped dressing tool would probably be a good thing to try. </p>
<p>I lathe turning stone scrapers are widely used. In this case a tough, abrasion resistant tool steel would probably be the best bet perhaps something like D2. </p>
<p>Similarly for turning it is probably well worth sawing the outside to a hex shape to get the section closer to round before you start. </p>
<p>Another approach would be to use a belt linisher with a coarse ceramic belt and a cylindrical grinding jig. </p>
<p>Of course if you just want a simple cylinder it would be logical to use core drills for both the inside and outside. In this case you might want to pre drill a pilot hole in the centre and fit a metal sleeve and use a plain pin instead of the pilot drill in the arbour. </p>
<p>Another big consideration is dust extraction to protect both the machine tool and your lungs a firebrick dust will wreak havoc on both. </p>
| 7964 | What's the most cost-effective cutter bit material that can be used to machine aluminum oxide? |
2016-03-16T14:46:13.220 | <p>Folks,</p>
<p>so today in the Lab I came up with this problem (see picture). It shows a cross section of a cutting machine. There's a blade moving with a constant speed $\frac{dx}{dt}$ towards the center of the roll made up of wood. This roll is turning with a adjustable speed $w$. Let's assume we keep $w$ also constant. So the questions I ask myself are as follows:</p>
<ol>
<li><p>How does the rotational speed w influence the thickness of the thickness of the fibers being cutted?</p></li>
<li><p>What is the mathematical function of the contact point (roll/blade)?</p></li>
<li><p>Do the fibers have a constant thickness?</p></li>
</ol>
<p><a href="https://i.stack.imgur.com/30UtU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/30UtU.jpg" alt="enter image description here"></a></p>
<p>Through careful consideration of the problem, I found the answers to 1) and 3).</p>
<p>Yes the thickness is influenced by $w$. For 3) it can be said that after one rotation of the roll, the fibers have a constant thickness. By observing the contact point on the wooden roll it can be seen that the point makes a spiral towards the center. Basically what is missing now, is the analytical/mechanical approach and mathematical formulation for this spiral.</p>
| |mechanical-engineering|materials| | <p>Your rotation speed $w$ is an <a href="https://en.wikipedia.org/wiki/Angular_frequency" rel="nofollow">angular frequency</a> (usually denoted with a small greek omega $\omega$, by the way). And its just what its name says: the change of angle of your roll in a certain amount of time. What is probably confusing is that the cutting velocity changes as the blade moves toward the rotation center of your roll. This is because the circumference depends on the radius.</p>
<p>The rotation period is $T=\frac{2 \pi}\omega$. The thickness of your fibers are defined by the distance the blade travels during one rotation period. With the velocity of your blade $\dot x$ this amounts to
$$
t= \dot x \cdot T = \frac{2 \pi \dot x}{\omega}
\quad\mbox{.}
$$
So as long as the ratio between angular velocity $\omega$ and blade velocity remains constant, the thickness is constant, too. This answers your questions 1 and 3.</p>
<p>The trajectory of the contact point is best calculated in a <a href="https://en.wikipedia.org/wiki/Cylindrical_coordinate_system" rel="nofollow">cylindrical coordinate system</a>:
$$
(r,\varphi) = (r_0-\dot x \cdot t,\omega \cdot t)
$$
with $r$ being the radial coordinate, $\varphi$ the angular coordinate and $r_0$ the outer radius of your roll. Here, $\varphi=0$ for $t=0$ <a href="https://en.wikipedia.org/wiki/Without_loss_of_generality" rel="nofollow">w.l.o.g.</a> If you need a different starting angle, just add e.g. a $\varphi_0$ to the $\varphi$ coordinate above.</p>
<p>To get cartesian coordinates, use the transformation
$$
(x,y) = (r \cdot \cos \varphi, r \cdot \sin \varphi)
\quad\mbox{.}
$$</p>
| 7975 | What is the analytical approach for a fiber cutting machine? |
2016-03-16T18:15:13.300 | <p>I'm trying to do some sheet metal work in Autodesk Fusion 360 and it has proved frustrating in the extreme. Maneuvering sketches and bodies to where I want them has proved extremely difficult. I understand that Autodesk Inventor has far better support for sheet metal parts, which is great if I could afford Inventor.</p>
<p>What is a workflow for making sheet metal parts in Fusion 360?</p>
| |cad|autodesk-inventor| | <p>This video about <a href="https://knowledge.autodesk.com/community/screencast/af415e28-f8ee-4b0d-b83b-b6076c2a941d" rel="nofollow">Faux Sheet Metal</a> parts was incredibly helpful.</p>
<ol>
<li>Start with a set of sketches that describe your sheet metal project.</li>
<li>Create patches based on those sketches. Each patch can then be moved to it's proper location.</li>
<li>After the patches are where they should be, then they can be extruded into bodies with the appropriate thickness. </li>
</ol>
<p>While this let's a user create parts that look like sheet metal parts, I believe there are significant differences between this process and a process that uses tools provided by Inventor or SolidWorks.</p>
| 7987 | Making sheet metal parts in Autodesk Fusion 360 |
2016-03-16T18:37:33.820 | <p>Shear areas have been obtained for a few different cross-sections. I-sections usually define their shear areas as equal to the product of the height of the section and the thickness of the web (assuming they fall under the thin-walled category).</p>
<p>However, what would the shear area for these sections be (assuming thin-walled theory applies)?</p>
<p><a href="https://i.stack.imgur.com/4DFiq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4DFiq.png" alt="enter image description here"></a></p>
<p>Should we adopt the smallest width which gives a continuous rectangle? But then the section to the right would have zero shear area.</p>
<p><a href="https://i.stack.imgur.com/qHcep.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qHcep.png" alt="enter image description here"></a></p>
<p>Or should we adopt the width which gives the largest continuous rectangle by area?</p>
<p><a href="https://i.stack.imgur.com/wckOr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wckOr.png" alt="enter image description here"></a></p>
<p>Or is there some other method?</p>
| |modeling| | <p>You can calculate the shear stiffness directly and then get an effective area after the fact if you'd like.</p>
<p>$$A_{shear}=\frac{V}{\gamma_{ave}\,G}$$</p>
<p>You can calculate average shear by averaging over the height (not area):</p>
<p>$$\gamma_{ave}=\frac1h \int \gamma\;dz$$</p>
<p>You can calculate shear strain from the shear stress:</p>
<p>$$\gamma=\frac{\tau}G$$</p>
<p>And shear stress from shear flow:</p>
<p>$$\tau=\frac{g}t$$</p>
<p>$$g(z)=\frac{V\,Q(z)}{I}$$</p>
<p>Now a lot of the variables cancel if you plug in at this point:</p>
<p>$$A_{shear}=\frac{h\,I}{\int \frac{Q}{t}\;dz}$$</p>
<p>Assuming a beam with shear center in the middle (for example symmetric):</p>
<p>$$A_{shear}=\frac{h\,\int^\frac{h}2_{-\frac{h}2}t\,z^2 dz}{\int^\frac{h}2_{-\frac{h}2} \frac{\int^\frac{h}2_zt\,z\; dz}{t}\;dz}$$</p>
<p>That will give you the effective shear area for any symmetric beam given the thickness/width as a function of z.</p>
| 7989 | Shear area of atypical section |
2016-03-16T22:57:03.320 | <p>I'm building a catapult and I asked a question and got <a href="https://engineering.stackexchange.com/a/7873/5412">this</a> response. Is there an equivalent for the wave spring but for angular motion like a torsion spring? Or any improved design on the torsion spring.</p>
<p><strong>Edit:</strong> An improved design by my opinion would be an increased spring constant in a decreased length</p>
| |springs|projectiles| | <p>The main advantage of wave springs compared to coil springs is that you can have the same stiffness and travel for less uncompressed length. This stems from the fact that a compression type coil spring needs to have an open helix to provide room for the spring to compress. </p>
<p>In comparison the various types of torsion springs don't change their dimensions much as they are loaded and so the advantages of a wave spring don't really apply. In fact their relatively compact size is one of their main advantages in a lot of applications. </p>
<p>What would constitute an improve design depends very much on exactly what the constraints of the system as a whole are. </p>
<p>There are a few types which fall broadly under the category of torsion spring, including solid bars, helical springs (as in mouse traps) and bundles of fibres. </p>
<p>There is a long history of using bundles of fibres as torsion springs in catapults dating back at least to the roman <a href="https://en.wikipedia.org/wiki/Ballista" rel="nofollow">balista</a> and medieval <a href="https://en.wikipedia.org/wiki/Mangonel" rel="nofollow">mangonel</a> </p>
| 7991 | Wave torsion spring? |
2016-03-17T04:39:32.493 | <p>I am trying to recreate the following model of a wooden roof truss provided to me by a truss manufacturer:
<a href="https://i.stack.imgur.com/ObODF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ObODF.jpg" alt="enter image description here"></a></p>
<p>My question is twofold:</p>
<ol>
<li><p>What is the common/correct way to model the boundary conditions and internal fixity conditions (pin pin?) for the elements as well as the structure itself? </p></li>
<li><p>Given generic roof loading in pounds per square foot, how should I distribute the loads to the truss? It seems to me that applying distributed live and dead loads for the top and bottom chords on truss elements poses a stability problem, as it was under my impression "idealized" trusses can only resist axial loading and thus loads can only be applied at the nodes. However this is a real truss, so I expect things may be different.</p></li>
</ol>
| |civil-engineering|structural-engineering|modeling| | <p>The most common way to model this structure is as follows (ignore the fact that the proportions are a bit off):</p>
<p><a href="https://i.stack.imgur.com/CHPnw.png" rel="noreferrer"><img src="https://i.stack.imgur.com/CHPnw.png" alt="enter image description here"></a></p>
<p>So, all of the diagonals are pinned-pinned. You'll notice in the schematic, however, that the chords (including the diagonals from the supports to the top chord) are not segmented at every point of intersection with the diagonals. Indeed, the chords are probably as long as possible, presenting splices at points E, J and M simply due to constructability demands (usually defined by the size of a standard truck). These splices between beams composing the chords, however, are usually designed so as to behave as a fixed joint, which is why I didn't put pins anywhere along the chords themselves.</p>
<p>Regarding the loads, you can just apply them as distributed loads. Since the individual member spans are quite small, almost no bending will occur and the vast majority of the internal forces will be axial regardless. Some analysis programs do allow you to apply distributed loads but set that they should be transformed into nodal forces, but there's really no need. And hell, the real truss <strong><em>will</em></strong> have that tiny bending, so you're more than welcome to include it in your analysis if you want.</p>
<p>The exception of course are the chords. Since they are not pinned at every joint, they will suffer significant bending, given that the diagonals will "deposit" their axial loads as concentrated nodal loads along the chords' effective spans.</p>
<p>And finally note that obviously the supports are different: one allows for horizontal displacement while the other doesn't.</p>
| 7998 | How do you model a real-life truss in structural analysis software? |
2016-03-17T07:45:42.193 | <p>I have thin weir, really just a sheet of metal with a smooth and straight upper edge. I want to know how much water will flow over this weir, when the water level on the higher side is a few cm above the edge.<br>
I'm interested in the special case when the water on the lower side i below the level of the weir:<br>
<a href="https://i.stack.imgur.com/mceEg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mceEg.png" alt="enter image description here"></a></p>
<ul>
<li>What formula describes this flow?</li>
<li>What are important caveats?</li>
</ul>
| |fluid-mechanics| | <p>Carlton is mistaken. Look at the exponent carefully. It should be 3/2, and not 2/3, this will obviously make a huge difference.</p>
<p>In <a href="https://engineering.stackexchange.com/a/22873/1832">another answer</a>, I give an explanation for why this equation is the case:</p>
<blockquote>
<p>At each location along the vertical distance $h$, the velocity is given by $\sqrt {2gh}$, where $g$ is the acceleration due to gravity, and this velocity of flow varies as $h$ changes along the vertical. So, you must integrate over the vertical distance since the relationship is nonlinear and you can't simply take the flow rate at the center. The first integral of the equation above is $\dfrac{2}{3}\sqrt{2g}\cdot h^{3/2}$.</p>
</blockquote>
<p>I neglect the C term which is experimentally derived.</p>
| 8001 | Formula to describe flow over thin weir |
2016-03-17T12:32:46.257 | <p>Display screens connected to cameras are significantly different from windows or mirrors. Looking through a window or mirror, we can focus on infinity, and we can move our vantage point to see a little different. With a display screen showing live camera footage, we see the field of view from a single point (the camera location) and our focus is on the screen.</p>
<p>Would it be possible to develop a screen + camera that is much closer in behaviour to a window or a mirror? I think the screen and camera would need to have the same surface area. Both would be sensitive to direction, and when camera pixel $(C_i, C_j)$ on the camera receives a photon with frequency $\nu$ from angles $(C_\phi, C_\theta)$, the screen would send a corresponding photon at frequency $\nu$ from position $(S_i, S_j)$ to direction $(S_\phi, S_\theta)$, where $(S_\phi, S_\theta)$ are calculated from $(C_\phi, C_\theta)$ mimicking either window- or mirror-like behaviour.</p>
<p>Is such a device theoretically possible? If yes, would such a device be technically feasible today? If yes, has there been any serious work on such devices? If it's theoretically possible but not feasible today, what would need to be developed before such devices may be on the horizon?</p>
<p>It should have a wide range of applications in <a href="https://en.wikipedia.org/wiki/Telepresence" rel="nofollow noreferrer">telepresence</a>, <a href="https://en.wikipedia.org/wiki/Augmented_reality" rel="nofollow noreferrer">augmented reality</a>, <a href="https://engineering.stackexchange.com/q/7965/3599">automotive engineering</a>, and surely a lot of other fields.</p>
| |materials|optics| | <p>Transmitting each individual <em>photon</em> is infeasible, given the amount of computation that would be required, but technology for capturing some information about incoming light direction already exists and is used in the <a href="https://illum.lytro.com/illum" rel="nofollow">Lytro</a> "light field" camera.</p>
<p>The corresponding light field display does not exist, as far as I'm aware. The Lytro system uses a conventional display with postprocessing that lets you adjust focal point, depth of field, etc. after the photo has been taken.</p>
| 8007 | Can cameras+display screens with window/mirror-like optical behaviour exist? |
2016-03-17T14:37:55.613 | <p>This question originates from <a href="https://engineering.stackexchange.com/questions/7955/truss-manufacturer-drawings-unclear">this</a> and <a href="https://engineering.stackexchange.com/questions/7998/real-life-truss-modelling">that</a> question. I have realized the concerns boil down to the following figure.</p>
<p><a href="https://i.stack.imgur.com/GDSFZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GDSFZ.jpg" alt="enter image description here" /></a></p>
<p>Simply put, is the truss above, with the displayed boundary/internal fixity conditions and distributed loading, structurally stable?</p>
| |structural-engineering|structures|structural-analysis| | <p>As mentioned in the comments to the question, since you are using a 3D analysis tool, you need to fix rotations around the X and Y axes. If you had been using a 2D analysis tool, simple pinned and roller supports would be suficient since rotations are only allowed around the Z axis, but since this is a 3D model, rotations around the other axes must be restrained.</p>
| 8014 | How Do You Determine Truss Stability in a Modeling Program? |
2016-03-17T14:41:26.263 | <p>What are the best materials and methods to keep the internal temperature of a device lower then the ambient temperature outside of the device? Based on the following information:</p>
<ul>
<li>Starting temp of device is 20°C</li>
<li>Ambient temperature around the device is 200°C</li>
<li>The time in this environment would be one hour (1hr), after which it would be removed</li>
<li>The electronics in the device would be very low powered</li>
<li>The size of the device would be 2" square</li>
<li>The device would be re-usable</li>
<li>The device is in an oven, hot air surrounding it</li>
<li>Inside the device is also air</li>
<li>It is a simple sensor outputting readings to a bluetooth module</li>
</ul>
<p>The goal would be to keep the internal components below 125°C</p>
<p>Any suggestions on how to achieve this would be helpful.</p>
| |electrical-engineering|thermal-insulation| | <p>There are drawer sized data safes available that are fire rated for 30 - 60 minutes. Perhaps you could inspect one of these for ideas on insulation /thermal mass. I think that they utilise some sort of cheap ceramic like concrete. Although 200 degrees isn't that much. I would have thought that some form of fire rated foam might work. It comes in spray cans like the common polyurethane foam filler stuff. You'd just need a suitable (metal?) container to fill.</p>
| 8016 | Suggestions on how to keep electronics cool in high ambient temperature location? |
2016-03-17T18:03:53.943 | <p>I want to know the heat conduction coefficient of a cup in order to investigate why tea cools at different rates in different containers. I know the cup is mostly plastic and paper. How do I determine the heat conduction? Do I use a known value?</p>
| |thermodynamics|heat-transfer|thermal-conduction| | <p>If you are interested in knowing why your tea is cooling in different times, you already answered the question yourself. It is due to the different thermal conductivities of the different items you use it to store.</p>
<p>$\dot{Q} = A U \Delta T \tag{1}$</p>
<p>is the heat flux you want to know. This gives you a relationship between the heat that is transfered to the enviroment and the dependance on the type of your container. So you need to know the thermal transmittance $U$. </p>
<p>If you are interested in calculating specific values for different items the following approach should do:</p>
<ol>
<li>Gather a list of all necessary heat transfer coefficients $\alpha_{ij}$ and thermal conductivity coefficients $\lambda_i$ In my opinion there is no need to calculate / measure these yourself, that's why books with tables of different coefficients exist. </li>
<li>Create a simple but accurate enough model of your container. A cup can be regarded as a hollow cylinder for example. Make a sketch of it!</li>
<li>Calculate the thermal transmittance $U$. Notice that different regions of your cup will have different heat transfer and thermal conductivity coefficients. For example the part of a coffee cup that is covered with a plastic lid and the lower part of the cup with no contact with the lid. Add them with the formulas for parallel and series (2). The attached image gives an example.</li>
<li>Calculate the overall heat flux. A higher heat flux means a quicker cooling time.</li>
</ol>
<p>$$U=\frac{1}{R_{overall}}$$</p>
<p>$$\begin{array}{c} \\
R_{overall} = \sum_{i=1}^n R_i &\text{in series} \tag{2}\\
\dfrac{1}{R_{overall}} = \sum_{i=1}^n \dfrac{1}{R_i} &\text{in parallel}
\end{array}$$</p>
<p>$R_i$ are the resistances for heat transfer and conductivity.</p>
<p>$R_{\lambda} = \frac{l}{\lambda A}$</p>
<p>$R_{\alpha} = \frac{1}{\alpha A}$</p>
<p>$l \equiv \text{length}$</p>
<p>$A \equiv \text{cross-sectional area}$</p>
<p><a href="https://i.stack.imgur.com/RwrGI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RwrGIm.png" alt="enter image description here"></a></p>
| 8024 | How can I determine the thermal conductivity of an open container? |
2016-03-18T06:54:59.630 | <p>I've seen two methods to describe porosity of porous media (specifically metal foams):</p>
<ol>
<li>Porosity $\epsilon$ which is a ratio between volume of voids and total volume of the medium.</li>
<li>PPI (Pores Per Inch) which is the number of pores in one linear inch.</li>
</ol>
<p>Unfortunately while reading some papers about the relationship between volumetric heat transfer coefficient and porosity in metal foams, some authors describe the range of porosity they used with $\epsilon$ while the others use PPI density, so is there any relation to connect both terms?</p>
<p>There is a similar question on <a href="http://www.cfd-online.com/Forums/fluent/86659-porosity-ppi-how-convert.html">cfd-online</a> and someone answered with this formula but I am not sure where it came from or if it's even correct:</p>
<p>$$\text{Pore density (ppcm)}=\frac{300}{D_{particle}}\sqrt{\frac{(1-\epsilon)^2}{\pi\epsilon}}$$</p>
| |mechanical-engineering|heat-transfer|porous-medium| | <p>The answer is system specific. That is, unless you know something else about the structure of your porous material, there is no clear relation between pore density and pore size. A simplified 2D example may look like this (where the circles are pores):</p>
<p><a href="https://i.stack.imgur.com/IGpUB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IGpUB.png" alt="same ppi different porosity"></a></p>
<p>In both of these samples the pore density in ppi is the same, but the porosity is different. They are approximately 4 ppi samples but in the one on the left the porosity is about 0.5 and the one on the right the porosity is much less than that.</p>
<p>I have confirmed this through a quick search of literature on foams like the one you describe. <a href="http://www.sciencedirect.com/science/article/pii/S0017931001002204" rel="nofollow noreferrer">This</a> one (behind a paywall) for example has a table where they list their samples and some of them have the same ppi but different porosities. In their works they describe their porous material as being made of pores with diameters $d_p$ and fibers of diameter $d_f$. Given that additional information, I guess someone could develop a relationship between pore density and pore size.</p>
| 8035 | Is there a relation between porosity and PPI density in porous media? |
2016-03-18T16:29:27.933 | <p>Consider a concrete column is under compression from top loading and also carries some shear stresses.</p>
<p>If take a plane 2d element in the column with these stresses and rotate it to the point it gives out the it maximum normal stress then rotate it to give out maximum shear stress which both values should be higher than our original calculated stresses.</p>
<p>Why don't we compare those value to the concrete's compression and shear strength instead?</p>
<p>I apologize if my question is too simplistic, I'm still a civil engineering freshman.</p>
| |mechanical-engineering|civil-engineering|structural-engineering| | <p>The short answer is because it is too complicated/impossible to do so.</p>
<p>Here is a diagram of the principal stress trajectories for an <em>uncracked</em> concrete beam under both flexure and compression:</p>
<p><a href="https://i.stack.imgur.com/8g7N8.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/8g7N8.gif" alt="Stress Trajectories"></a></p>
<p>As you can see the orientation and magnitude of the principal stresses will change depending on the point you are interested in and the applied loads. We know that concrete is weak in tension. So if we are looking at a location of principal tensile stress we can compare this to the tensile capacity of the concrete (which is often considered to be a function of $\sqrt{f_c'}$). </p>
<p>What if the principal tensile stress exceeds the tensile capacity of the concrete? </p>
<p>Well at that point the concrete may fail. But this doesn't mean the whole element will fail. It means that it will <em>crack</em> at that location. But that is OK, that is what reinforcement is for!</p>
<p>So now we have a concrete element with a crack (or many cracks!), and reinforcement to hold the pieces together:</p>
<p><a href="https://i.stack.imgur.com/GFxBn.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/GFxBn.gif" alt="TensionCracks"></a></p>
<p>If we now want to calculate our principal stresses, what is the state of stress at a particular point? We have some stress being carried by the reinforcement, some stress being carried by aggregate interlock along the cracks, some being carried by compression, and some voids where no stress can exist - how much goes into each mechanism? We can't simply use formulas like $\nu = \frac{VQ}{It}$ since this only applies to a uniform material.</p>
<p><em>We can't determine the state of stress with any reasonable certainty in a cracked concrete section$_1$.</em></p>
<p>So what can we do now? Well, we do lots, and lots of <a href="https://www.concrete.org/publications/internationalconcreteabstractsportal.aspx?m=details&ID=7920" rel="noreferrer">tests</a> and then fit a design equation to the results.</p>
<p>You mentioned columns in your question. Columns are dominated by compressive stresses, so cracking is often not as much of an issue. However, there are still complicating factors which will make it difficult/impossible to determine the stress state. In fact, the commentary of <a href="https://www.concrete.org/store/productdetail.aspx?ItemID=31814" rel="noreferrer">ACI 318</a> says: </p>
<blockquote>
<p>The actual distribution of concrete compressive stress is complex and usually not known explicitly. ... The Code permits any particular stress distribution to be assumed in design if shown to result in predictions of ultimate strength in reasonable agreement with the results of comprehensive tests.</p>
</blockquote>
<p>So again, we are forced to take the easier route of assuming a simplified stress state and confirming that is safe according to tests. </p>
<p>The uncertainty related to using these simplifications is incorporated into the safety factors used in building codes.</p>
<p>It would be much more satisfying to have a design methodology which is built on the principal stresses. This has apparently been tried in the past, but was always unsuccessful due to the difficultly in determining the stress state$_2$. </p>
<ol>
<li><p>Kong, F. K., & Evans, R. H. (2013). Reinforced and prestressed concrete. Springer.</p></li>
<li><p><a href="https://www.concrete.org/publications/internationalconcreteabstractsportal.aspx?m=details&ID=7920" rel="noreferrer">ACI-ASCE Committee 326 (1962). Shear and Diagonal Tension</a> </p></li>
</ol>
| 8043 | Why don't we use principal maximum stresses when designing a structural element? |
2016-03-18T17:35:16.133 | <p><a href="https://i.stack.imgur.com/aj7XI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aj7XI.jpg" alt="Increase yield strength"></a></p>
<p>I know that with strain-hardening you increase the yield and tensile strength. So if you unload at point D, then you also have cold worked it a bit right?
So the stress-strain curve will look different after point D and have a higher tensile point?</p>
| |materials|metallurgy| | <p><strong>Summary: If you are loading and unloading the same sample, the stress-strain curve will not change and the tensile strength will be the same, because stress and strain are calculated from the initial geometry. If you are comparing two otherwise-identical samples with different levels of cold work or strain hardening, their stress-strain curves will be different and they will have different tensile strengths.</strong></p>
<p>If you unload at D, then retest the sample (assuming necking has not yet occurred) it will effectively pick up where it left off. The stress-strain curve will follow the right-hand elastic curve (straight line) portion of the diagram back up to point D, and will then continue deforming plastically.</p>
<p>Put another way, by unloading and then starting the test over you shift the origin of the plot to the right, to where the end of the unload line is. Note that from the new origin, D becomes the yield point, which is higher than before, and the elongation at failure is reduced because it is closer to the origin.</p>
<p>If you retest the same sample, the tensile strength does not increase. If you instead compare two samples of the same initial geometry and material, except one is as-annealed and the other is cold worked, the latter will have higher tensile strength. The reason for the discrepancy between the two scenarios (retesting vs two differently worked samples) is that plastic deformation has changed the geometry in the former scenario.</p>
<p>To properly reflect the change in material properties when retesting would require measuring the narrower, longer sample and computing the new stress and strain curves from the new values after unloading. If that were done, the tensile strength recorded for the retested and remeasured sample would be higher than the initial test (stress increases as area decreases). Using the original geometric measurements would cause the sample to continue following the same stress strain curve. Another way of thinking about it is that retesting the same sample is like comparing two samples with different cold work <em>and</em> different geometries, but basing the calculations on only one samples geometry, in such a way that the curves are identical.</p>
| 8045 | Does unloading beyond yield point also affect tensile strength? |
2016-03-18T20:33:18.630 | <p>Using DePriester Chart and Given one of mole fractions ($z$), pressure and temperature we can acquire K-values for that properties and bubble and dew properties.</p>
<p>Now there are sometimes that we get some K's and using the summation of $y/k$ we get some quantities that are not 1, like 0.9 or 1.2. My question is about those values, are they showing some physical property?</p>
<p>If I wanted to calculate the dew point pressure at a given temperature and I take arbitrary pressures (including the dew point pressure) I get the following results.</p>
<pre><code>[p] = psi sum of y/k
- - - - - - - - - - - -
100 0.828
126 1.000
150 1.174
</code></pre>
<p>Do the values for the pressures above and below the dew point pressure have a physical meaning?</p>
<p>For example, showing us that we are in two-phase, SH vapor or SC liquid region?
Or are they just showing there isn't VLE with that given property?</p>
| |thermodynamics| | <p>You can represent K as $K_i = \frac{p_i^{sat}}{p}$ for an ideal mixture using Raoult's law. </p>
<p>This equation directly yields the correspondance of $K$ and $p$. If the pressure is below the dew point pressure $K_i$ will be larger. Hence $\frac{z_i}{K_i}$ will be smaller and subsequently the sum over it.</p>
<p>What this tells you is</p>
<p>$p$ yields $\sum \frac{z_i}{K_i} < 1$ pressure is below dew point pressure</p>
<p>$p$ yields $\sum \frac{z_i}{K_i} > 1$ pressure is above dew point pressure</p>
<p>Note: This is for a fixed temperature.</p>
<p>Alternatively you can make the same deductions with $K_i = \frac{y_i}{x_i}$ and the knowledge that at dew point the first drop of liquid forms.</p>
| 8047 | Physical meaning of some earned values using DePriester Chart |
2016-03-18T21:38:23.820 | <p>I have a situation where a 5mm steel rod needs to be in contact and rolls over segments of a 80mm steel pipe's inner face. Would it be a good idea to loosely slip a Teflon bushing over the rod and use it as a roller? I haven't seen loose fitted bushings, so I'm not sure if it'd be right to let a bushing rotate like that. </p>
| |mechanical-engineering|bearings| | <p>This is fine in principal, there are a few applications where a similar arrangement is used, in fact it is common for nylon wheeled castors to just have a plain bore. </p>
<p>In this situation the bushing will tend to travel along the rod as it rotates unless it is constrained so you will need something to locate it axially. </p>
| 8048 | Using a Teflon bushing as a roller |
2016-03-18T22:52:11.110 | <p>I'm curious about whether a typical liquid CPU cooler (like, e.g., this <a href="http://rads.stackoverflow.com/amzn/click/B00A0HZMGA" rel="nofollow">Corsair H60</a>) would be able to effectively cool a square sheet of aluminum with 15" sides and a thickness of 1/8".</p>
<p>I'm not even sure how to properly formulate the question so I'll describe the envisioned application. I'd attach the cooling plate to the bottom of the aluminum sheet and build a case for it with a cut-out for the fan. The result would (hopefully) be a refrigerated cooling pad for my constantly overheating laptop.</p>
<p>Is such an idea feasible? How powerful of a cooler would be needed to effectively chill such a sheet of aluminum?</p>
<p>Assume an ambient room temperature of 70 degrees Fahrenheit and that the bottom of the laptop is roughly 160 degrees Fahrenheit. I'd like to reduce the temperature of the bottom of the laptop to 120 degrees Fahrenheit. The bottom of the laptop is aluminum and about 15"x15". There is a gap between the aluminum sheet and the laptop bottom of about 1/8".</p>
| |mechanical-engineering|cooling| | <p>Your best bet may be to make a hole in the aluminium sheet and mount the fan to either blow or suck the air from between the laptop base and the aluminium sheet ie using the 1/8" gap...</p>
| 8049 | How effectively would a typical liquid CPU cooler cool a 15"x15"x1/8" sheet of aluminum? |
2016-03-19T08:23:35.063 | <p>I found this strange looking bit in a bit set of uncommon bits, including Security Torx, hexalobular, tri-wing, spanner head, 12-point flange, Torq-set and others. It's the only one that I could not identify, what is it for? </p>
<p><a href="https://i.stack.imgur.com/6A1Be.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6A1Be.jpg" alt=""></a></p>
<p><a href="https://i.stack.imgur.com/SmZP6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SmZP6.jpg" alt=""></a></p>
| |mechanical-engineering|fasteners| | <p>There is one sold on Amazon.com, an "Eazypower 80935 1/4 inch Y-Shape Cup Hook Installer and Remover". <a href="https://rads.stackoverflow.com/amzn/click/com/B000PSABL0" rel="nofollow noreferrer" rel="nofollow noreferrer">https://www.amazon.com/Eazypower-80935-4inch-Installer-Remover/dp/B000PSABL0</a></p>
<p>I usually use a pair of pliers to screw hooks in or out. Quite honestly, I like the wing nut driver idea better.</p>
| 8054 | What is this Y-shaped screwdriver bit, and what is its purpose? |
2016-03-19T15:52:00.510 | <p>This question is so fundamentally basic that I am almost embarrassed to ask but it came up at work the other day and and nearly no one in the office could give me a good answer. I was calculating the shear stress in a member using the equation, $\frac{Tr}{J_T}$ and noticed, that for a shaft with a circular cross section, $J_T = I_P$.</p>
<p>Both $I_P$ and $J_T$ are used to describe an object's ability to resist torsion. $I_P$ is defined as, $ \int_{A} \rho^2 dA $ where $\rho$ = the radial distance to the axis about which $I_P$ is being calculated. But $J_T$ has no exact analytical equations and is calculated largely with approximate equations that no reference I looked at really elaborated on.</p>
<p>So my question is, what is the difference between the Polar Moment of Inertia, $ I_P $, and the torsional constant, $ J_T $? Not only mathematically, but practically. What physical or geometric property is each a representation of? Why is $J_T$ so hard to calculate?</p>
| |mechanical-engineering|structural-engineering|applied-mechanics|stresses| | <p>This is almost a coincidence, and it is only true for solid or hollow circular cross sections. Of course shafts carrying torsion often <em>are</em> circular, for reasons that are independent of the question! </p>
<p>The torsion of a circular shaft is physically simple because of the symmetry of the circular shape. By symmetry, the stresses and strains at any point can only be a function of the radial distance from the centre line of the shaft. By Pythagoras' theorem, you can take an arbitrary pair of axes and express the radius as $r^2 = x^2 + y^2$. </p>
<p>Using that fact, you can convert the integral over the cross section into the sum of two integrals in the $x$ and $y$ directions, and again by symmetry those two integrals must be equal to each other. </p>
<p>The form of the integrals happen to be exactly the same mathematical form as for the second moments of area of a circular beam, which leads to the result you asked about.</p>
<p>This doesn't work for non-circular sections, because the stress distribution is not radially symmetrical. For example if you compare the torsion constant and polar moment of a solid square section, you will find the "constants" in the two formulas are different. The more the cross section deviates from a circle, the bigger the difference will be.</p>
<p>The torsion constant for an complex shaped section (for example an I-beam) is hard to calculate because the stress distribution over the section is complicated, and there is no simple "formula" for it that you an integrate mathematically. Many of the formulas for torsion in engineering handbooks are based on simplified assumptions rather than "exact" mathematical solutions. </p>
<p>But in real life the "errors" are not too important, because when a torsional load is applied to a non-circular structure, the cross sections "warp", i.e. <em>they no longer remain plane</em>. In real life, the amount of warping is often unknown, because the restraints at the ends of the shaft affect it. If you really need an accurate estimate of the torsional stiffness of a non-circular component, you have to make a full 3-D model of the component itself and how it is fixed to the rest of the structure. If you make a model with that level of detail, there is not much point in reducing the answer to one number just so you can call it "the torsional stiffness".</p>
| 8064 | What is the difference between the Polar Moment of Inertia, $ I_P $ and the torsional constant, $ J_T $ of a cross section? |
2016-03-20T04:18:43.430 | <p>I'm trying to use a 1500 RPM-rated electric motor to accelerate a (light, 25kg) shaft up to 50 000 RPM. At these speeds I figure using a belt system or roller chains is out of the question due to wear and friction; are these kinds of speeds possible to achieve reliably with a mechanical transmission or do I need to turn to hydraulic transmission? Or what is an efficient way of achieving this?</p>
| |gears|energy| | <p>It sounds like you need something like a <a href="http://www.cutawaycreations.com/cutaways.php?project=pratt-whitney-pt6-turboprop-engine#/IMAGES/pratt-whitney-pt6-turboprop-engine/images/image4.jpg" rel="nofollow noreferrer">turboprop transmission</a> operating in reverse. These are generally planetary gearboxes for good balance at high speeds.</p>
<p><a href="https://i.stack.imgur.com/CWNvN.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CWNvN.jpg" alt="enter image description here"></a></p>
| 8074 | Getting high RPM out of a low RPM motor |
2016-03-20T16:18:43.787 | <p>I have an outer pipe and an inner pipe a lot smaller than the inside of the the outer pipe, and the head of the (mini) turbine will be connected to a metal rod. There will be wires connected to the head and it will be turning to the wind. How should I set up the wires?</p>
| |design| | <p>A <a href="https://en.wikipedia.org/wiki/Slip_ring" rel="nofollow">slip ring</a> connector is the usual way to solve this problem. The electrical contact is made by fixed brushes which make sliding contact with a rotating ring. </p>
<p>The key point is to ensure that you maintain a good and consistent electrical contact. It is common to use brass slip rings and graphite contactors held in place with springs and designed to wear and be replaced periodically. </p>
<p>Graphite has the advantage that it combines good electrical conductivity with self lubrication and is soft enough to wear to conform precisely to the shape of the surface of the slip ring. </p>
<p>It is also important that the enclosure is designed to prevent contamination of the contact surface by dust, oil, moisture etc. </p>
| 8080 | How to make a wire turn with a turbine head |
2016-03-21T04:43:23.813 | <p>First a little background, I've been working in Michigan as an engineer since getting my BSME in 1998 (I graduated with my MSME in 2010). I've worked for a number of companies in a variety of industries as an engineer, but never under the direct supervision of a licensed PE since all of the work I've done has been considered under "industrial exemption". </p>
<p>Now however, I'm becoming less interested in simply being an employee and more interested in working for myself and being a PE would be advantageous. I'm reasonably sure I can pass the exam (with copious amounts of studying, naturally), but my concern is getting my experience verified. The PE's that I know, I've never worked with, either as a subordinate or as a peer. </p>
<p>Has anyone had difficulty finding enough PE's to verify their experience when getting licensed? If so, were you successful in getting your experience verified? How do you recommend finding PE's to verify engineering work experience? </p>
| |licensure| | <blockquote>
<p>How can I verify my engineering work experience without having worked under a PE?</p>
</blockquote>
<p>Many states offer the ability to substitute sufficient / substantial work experience for directly supervised work experience. The key difference here is that the obligation is upon the applicant to demonstrate how the in lieu of work still meets or preferably exceeds the expectations of the professional board with regards to the quality of the work produced. In other words, because you can't produce a PE to vouch for direct supervision then you have to demonstrate why the work ought to qualify.</p>
<p>From what I can tell of <a href="http://www.michigan.gov/lara/0,4601,7-154-72600_72602_72731_72865_73185-141914--,00.html" rel="nofollow noreferrer">Michigan's work experience requirements</a>, you ought to be able to substitute sufficient work experience.</p>
<blockquote>
<p><strong>Work Experience Requirements</strong><br />
All applicants must provide verification of at least 4 years of acceptable engineering work experience obtained after having received an acceptable bachelors degree. Work experience must be verified by five persons, three of whom must be licensed professional engineers.</p>
</blockquote>
<p>The devil is obviously in the details, and it would behoove you to contact that board for exact details. But at first glance, it would appear that you'll be fine.</p>
<hr />
<p>To be a little more prescriptive in what you specifically need to do:</p>
<ol>
<li><p>Contact the PE's that you've worked with before. Find as many that you can that are willing to vouch for the quality of your work. It's important to keep in mind that they are vouching more for your character and their perception of you following a particular process as opposed to direct supervision. Some states call these "character reference" recommendations.</p>
<p>Michigan expects at least 3 PE's to vouch for your work, you would be better off getting more than that if you can. There's no harm in submitting more than 5 verifications as well, especially if they all hold PE's.</p>
</li>
<li><p>Start documenting the work experience that you've already completed. Some states count a Masters of Engineering as an equivalent to one year of work experience. In that documentation, call out the processes that you followed and make sure to note the review portion of those processes. The state board is interested in work that demonstrates the full engineering process showing the beginning of the work through till the end and including the review to potentially improve the process itself.</p>
</li>
<li><p>Finally, really start studying the law and regulations surrounding the particular field of engineering that you wish to advertise your services within. And let me be straight-up brutally honest for a moment: Sealing a document merely means that you're willing to accept liability for a project should something that was reasonably foreseen go wrong. That means you can personally be civilly or criminally held liable should there be a significant problem on a project. Make sure you're comfortable with reasonable standards of care for the projects you want to engage so you can demonstrate due diligence in your offerings.</p>
</li>
</ol>
| 8090 | How can I verify my engineering work experience without having worked under a PE? |
2016-03-21T17:47:11.380 | <p>I have 8" thick reinforced concrete walls and I would like to anchor my horizontal corner reinforcement with only 1 hook (I have only 1 layer of reinforcement in both walls). The problem is CSA A23.3 (same for ACI if I'm not mistaken) restricts minimum hook length to 6" or $8d_b$ = 120mm = 5" in my case. </p>
<p>If I want to put my reinforcement in the middle of my wall, there is no way for me to get that 6 inches. Can I simply assume that the bar is held by my vertical reinforcement (see picture)? </p>
<p><a href="https://i.stack.imgur.com/o39uF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o39uF.png" alt="CORNER"></a></p>
| |reinforced-concrete| | <p>I'd agree, you cannot fit a <em>horizontally oriented</em> 5" 90° bar bend in an 8" thick wall if you want your rebar curtain to be located at the center-line of the wall. However, there is no reason that the bent bars will not fit if you orient the bend vertically. The bars can be placed such that the 5" 90° bar bend is oriented vertically.</p>
| 8098 | Basic Hook development length using CSA A23.3 provisions |
2016-03-22T11:45:11.853 | <p>How can the microstructure of steel after welding be determined in modern computer software?</p>
<p>I am aware that there are 'classic' methods of predicting microstructures like Schaeffler, DeLong or WRC constitutional diagrams. What I am wondering about is: which algorithms are used in programs based on finite element method (FEM) computing (i.e. ESI Group software, particularly SYSWELD)</p>
<p>What I want to find out in particular:</p>
<p><strong>1. How does a FEM model made of austenite differ from the one that is made of ferrite?</strong></p>
<p><strong>2. Is a FEM model enough to determine the microstructure of a material, that is represented by that model? If not, what else do you need to examine transformations of microstructure during the welding process?</strong></p>
| |metallurgy|finite-element-method|structural-analysis|software|welds| | <p><strong>Summary:</strong></p>
<p><strong>1) The answer to this question is difficult. You would need to know how austenite and ferrite behave in relation to what you are doing to them. You would also need to know their compositions, temperature field, etc. The results here could vary significantly depending on the specific parameters and how they change with time and with each other.</strong></p>
<p><strong>2) Yes and no. You can get a statistical model of each FEM element at a macro scale to determine the statistical nature of the microstructure such as grain size, particle density, etc. Or you can model what happens at a microscale when welding to determine what the microstructure might look like under a microscope. To do either requires a lot of detailed thermodynamic data about all of the components present and their phases, as discussed below.</strong></p>
<h1>More Specifically...</h1>
<p>My approach to the problem would be to use statistical modeling to determine grain size, particle size and distribution, etc. The data would come from phase diagram data based on known compositions, together with assumed kinetic governing equations. The actual thermal and compositional kinetics are governed at a bulk scale by well known PDE models, but certain features such as size and distribution are determined by microscale kinetics. What phases appear and the order in which they appear are governed by the phase diagrams. We can generally assume that bulk kinetics provides an input into the microscale kinetic and thermodynamic relationships, but that the reverse is largely irrelevant. The microstructural data in turn provides an input into structure-property relationships.</p>
<p>A complete model would tie all of these together more-or-less in that order, and would look something like:</p>
<ul>
<li>Use the bulk scale kinetics to get the temperature and composition profiles for each element.</li>
<li>Use the temperature and composition profiles to generate statistical microstructural data for each element.</li>
<li>Use the statistical microstructural data to determine properties.</li>
</ul>
<p>The algorithms for common software are generally proprietary, so I can't say for sure, but I believe any package such as SYSWELD or MAGMA uses something along these lines. SYSWELD may even use the Schaeffler, DeLong, and WRC diagrams as part of its thermodynamic modeling. It depends to what degree they are making assumptions about the thermodynamic data and how much effort they've put into that part of their model.</p>
<h1>Microstructural FEM</h1>
<p>FEM may be used to model microstructural behavior at a microscale, such as mechanical and thermal behavior. Generally this is done by capturing a microstructural image representation, either by microscope (optical or SEM) for 2D or by computed x-ray tomography (CT) for 3D, and converting the representation into an FEM by identifying or segmenting different phases and their interfaces, and assigning appropriate (often anisotropic) material properties to each phase and each phase-pair interface.</p>
<p>To do all this, you need to be able to accurately segment phases, interfaces, and crystallographic orientation, which may take some expensive equipment and characterization work. Alternately, a phase field model may be used to attempt prediction of the microstructural morphology, and then relevant data captured from the phase field model. There are limitations in using a phase field model this way, which are discussed in the next section.</p>
<p>There is a tool on <a href="https://nanohub.org" rel="noreferrer">nanoHUB.org</a> which does microstructural FEM for 2D images called <a href="https://nanohub.org/tools/oof2" rel="noreferrer">OOF2</a>. To use the tool you would need to create an account, and it is generally intended for educational purposes, but should give a general idea how a microstructural FEA might work in the 2D case. You might need to upload your own microstructure image, it's been awhile since I've used it and I've forgotten the details.</p>
<p>The results of microstructural FEM are useful for determining how textured microstructures might behave. They are also useful for determining how microstructure can be linked to fatigue properties by identifying stress concentrations in the microstructure and how the microstructure might play a role in crack and void initiation.</p>
<h1>Phase Field Models</h1>
<p>To model phase transformation kinetics at a microscale, generally phase field models (<a href="https://en.wikipedia.org/wiki/Phase_field_models" rel="noreferrer">Wikipedia</a>) are used. The models involve a number of moving parts, so to speak, but are sometimes faster and usually more robust for capturing moving interfaces than with traditional FEM models.</p>
<p>The primary concept is that, for a field of elements containing two phases, the phase discrimintation of the entire field may be modeled with a scalar value varying from 0 to 1. If the value of an element is (very close to) 0 it is one phase, and (very close to) 1 the other phase. If it has an intermediate value it is part of an interface between the phases. Thus rather than a sharp interface as would be required with an FEM, the interface is modeled by assuming it is diffuse.</p>
<p>Phase field models typically also track composition, temperature and free energy, and have a collection of governing equations which must be solved at each time step to determine the evolution of the next step. To use a phase field model thus requires knowledge of temperature dependent free energy curves and temperature dependent diffusion rates, both thermal and compositional, among other things.</p>
<p>It is possible to model diverse microstructural evolution phenomena with phase field models, including:</p>
<ul>
<li>planar to dendritic solidification (<a href="https://nanohub.org/tools/vkmlggs" rel="noreferrer">nanoHub.org tool</a> and <a href="http://solidification.mechanical.illinois.edu/Movies/Plapp/dendlowanis.webm" rel="noreferrer">solidification.org video</a>)</li>
<li>unstable cellular eutectic solidification (<a href="http://solidification.mechanical.illinois.edu/Movies/Plapp/Eutectic_Colonies.webm" rel="noreferrer">solidification.org video</a>)</li>
<li>spinodal decomposition (<a href="https://nanohub.org/tools/vkmlsd" rel="noreferrer">nanoHub.org tool</a>)</li>
<li>grain growth (<a href="https://nanohub.org/tools/vkmlpsgg" rel="noreferrer">nanoHUB.org tool</a>).</li>
</ul>
<p>and certainly much more is possible, if challenging.</p>
<p>I am unaware of any professional phase field modeling packages, as there are drawbacks limiting their usefulness outside academia and early-stage research. One limitation is that, depending on the model, the specific variables may not have a clear relationship with physical values that can be experimentally determined. Thus, sometimes the model parameters need to be adjusted systematically until the results "look right." Additionally, getting useful information out of the model is another issue due to the same discrepancies between model parameter values and physical values. It is also possible to produce non-physical results quite easily without careful tailoring of governing equations to the specific model. Validation is another issue as performing the experiments is at best an expensive and time consuming process, and at worst virtually impossible depending on the specific parameters involved. Research is of course focused on reducing these issues, but because phase field models are relatively new (~10 years old at the earliest) there is much work yet to be done.</p>
<p>Generally phase field models are mostly useful for drawing a pretty picture of what a microstructure might look like without performing expensive experimentation and microscopic examination. They are also useful for creating animations of microstructural evolution. In the future their use may expand to predicting features such as statistical modeling and capturing data for FEM, but the limitations above restrict those uses.</p>
<h1>Statistical Thermodynamic Models</h1>
<p>Generally, at a useful level, most engineers deal with bulk properties. After all, most engineers are designing products at a visible, macroscopic scale. As a result, the specifics of a tiny fraction of the microstructure isn't particularly useful directly to how the product might behave at a macroscopic level. Instead, we want to determine how the product behaves as a bulk.</p>
<p>To model the bulk microstructural evolution and final properties of a material is usually done with a CALPHAD (Calculation of Phase Diagrams, <a href="https://en.wikipedia.org/wiki/CALPHAD" rel="noreferrer">Wikipedia</a>) program. Microstructural CALPHAD models don't generate pretty pictures like in the previous two sections, but instead generate a statistical representation of certain classes of microstructure by generating a grain size or particle size and density distribution based on thermodynamic, kinetic, compositional, and temperature data.</p>
<p>Such a model can be used in conjunction with a process FEM to determine local microstructural distributions for each finite element. <a href="http://www.thermocalc.com/" rel="noreferrer">Thermocalc</a> does thermodynamical statistical modeling and is a CALPHAD program. <a href="http://www.magmasoft.com/en/" rel="noreferrer">MAGMA</a> casting process simulation software combines statistical thermodynamic modeling with FEM in some of their alloy packages. A MAGMA user might then be able to predict some statistical data throughout the bulk of their product, and then generate scalar fields representing mechanical properties which vary over the product. It appears SYSWELD does the same thing for heat treatment and the welding process, probably by the general method described here.</p>
<h1>References</h1>
<p><a href="https://nanohub.org/" rel="noreferrer">nanoHub.org</a> - A site with many computational tools and educational resources focused on nano-scale. Some information and tools are related to larger scale modeling, especially OOF2 (Object Oriented Finite Element Analysis, 2D) and VKML (Virtual Kinetics of Materials Laboratory) tools.</p>
<p><a href="http://www.solidification.org/" rel="noreferrer">solidification.org</a> - A site with a number of neat movies of solidification processes, both experimental and phase field simulation.</p>
<p>I do not endorse any linked sites or softwares, the links are only intended for referential and educational purposes.</p>
| 8111 | Determining the microstructure of steel after welding in modern computer software |
2016-03-22T15:36:35.213 | <p>I'm constructing 2.2 meters span, 1 m wide shelf. Want to make it cheap but thin.
I'm planning to use 18mm thick OSB board and reinforce it with a layer of stronger material stuck to it's bottom.</p>
<p>My first idea for reinforcement is 2 strips of 60*4 mm steel.
I'm not very familiar with the subject. I found OSB modulus of elasticity <a href="http://article.sapub.org/10.5923.j.cmaterials.20140402.03.html" rel="nofollow noreferrer">here</a> , transformed width of the OSB board to it's steel equivalent, that gave me 38mm. Then I ran it through L-beam moment of inertia online calculator, and beam calculator assured me that the sag would be much less than my acceptable 1cm. (The load is under 100kg)</p>
<p>My second idea is to use glass fabric instead of steel. Glass itself is just 3 times less strong than steel, so that makes this option at least worth evaluating. </p>
<p>Hence comes <strong>the question</strong>:
Where do I find properties of the layer I'll get when I put fiber glass and pour epoxy on it? The only thing that's stated for the "structural fiberglass fabric" available on the market is it's ultimate tensile strength.</p>
<p>I'm talking about the glass fabric, not the plastic sheet reinforced with it:
<a href="https://i.stack.imgur.com/1BCei.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1BCei.jpg" alt="enter image description here"></a></p>
| |materials|structural-analysis|composite| | <p>Hi to the next person to read this,</p>
<p>The preceding appears to be all correct, however, it appears to be for a solid composite beam. If you are using a USB board as a core, you must rather do composite beam calculations.</p>
<p>I landed on this post because I wanted to refresh my memory to calculate composite beams for a snow yatch I am building (like an iceboat but on special skis I have laminated). The cross beam is 12 feet, laminated with 6oz unidirectional ( I presume 140GPA, thanks wwarniner!) and woven 6oz 45 degree twill for torsional rigidity and capping (this is more complex with poisson ratios and that stuff (?). Anyway, I have 400 lbs centre load, simply supported and I will pre-stress for extra stiffness (laminate tensile side flat, then compression side with twice the curvature desired. Loads up the core in compression and adds loads of Stiffness! I am looking to know roughly how many layers Of UD I will need.</p>
<p>The process is as follows:</p>
<ol>
<li><p>You calculate the total moment of inertia for each rectangular area consdering the distance from the neutral axis that really pays off!!! That is why surface layers in a sandwich make it so light and stiff.</p></li>
<li><p>You then calculate the contribution of each area multiplied by it's stiffness (E) for the flex. Soft materials and layers near the neutral axis contribute insignificantly, as opposed to surface layers, particularly UD in my case. Note: A honeycomb or foam core as you can imagine contributes insignificantly, all the work is done by the extreme fibres. However, you must be concerned about core shear in this case)</p></li>
<li><p>You can do the same for strength using UFM or UTS (ultimate tensile strength)</p></li>
</ol>
<p>Reminders: Rectangle I = b*h^3/12, Z = I/c. (C is distance to extreme fibre, that is h/2). Right? (This is all from memory, so pls chk 2 b sure)</p>
<p>That is what I remembered from the Gibbs and Cox composite design bible I used many years ago. Once you use it, you ré member the principles for life ;-)</p>
<p>BTW, if anyone would like the most delightful engineering read ever, I strongly recommend a book I read about the development of all this in war times, "The Science of Strong Materials" by a British engineer by the name of J.E.Gordon. Composite calculations came from plywood calculation for aircraft) Bloody good read if I do say so meeself ;-)</p>
<p>Cheers!
Paul Isabelle,
Industrial designer and tinkerer</p>
| 8118 | How do I calculate the structural properties of a fiberglass laminated wood panel? |
2016-03-22T16:09:16.050 | <p>I have modeled a beam as fixed-fixed. The reaction forces are -1494 lb.ft and 761 lb (vertical). There is no horizontal reaction.</p>
<p><a href="https://i.stack.imgur.com/drvLH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/drvLH.jpg" alt="enter image description here"></a></p>
<p>Now I know this beam is welded to another element. I am trying to design the weld for the necessary forces. How do I transfer the reaction forces at the fixed end to the weld? Should I invert the directions of the reactions or should I preserve the same directions? </p>
| |structural-engineering|structures|structural-analysis| | <blockquote>
<p>Should I invert the directions of the reactions or should I preserve the same directions?</p>
</blockquote>
<p>As long as you keep your directions consistent while solving for the weld forces, it won't matter whether you have them positive or negative. Referring to my answer to your <a href="https://engineering.stackexchange.com/a/8104/2246">first question</a> on weld strength, the $f_{xw}, f_{yw}, f_{zw}$ terms are all squared when solving for $f_w$, so directionality doesn't matter so long as you keep it consistent.</p>
| 8119 | How to transfer reaction forces of beam to a weld? |
2016-03-22T20:38:47.533 | <p>I have a vehicle with 4 wheels. Say I replace the wheels with large helical gears (pinions). If I set the vehicle on linear helical racks, it should be able to travel in one dimension. Say I replace the racks with a similarly toothed ground surface, could the vehicle travel in 2 dimensions (without rotating the body of the vehical)? I think it would be possible with opposing front and rear wheel drives. It should operate like a "screw" in the second dimension. Would the ideal helix angle be 45 degrees (to support movement in both directions)? How will speed, Torque, and service life be affected?</p>
<p><a href="https://i.stack.imgur.com/17qHL.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/17qHL.gif" alt="enter image description here"></a></p>
| |mechanical-engineering|gears| | <p>Is this theoretical. If so I will bow out, but if not, Airtrax Sidewinder Forklift already does this. See the video. <a href="https://www.youtube.com/watch?v=vAiwLRGsNrE" rel="nofollow">https://www.youtube.com/watch?v=vAiwLRGsNrE</a></p>
<p>techtooler</p>
| 8127 | Multi-Directional Rack and Pinion Drive - Vehicle Motion Using Helical Wheels |
2016-03-22T21:56:13.437 | <p>I am trying to solve a fairly simple question but I'm kind of stuck on the technicalities:</p>
<blockquote>
<p>Imagine that water is evaporating into initially dry air in the closed
vessel shown schematically in Fig. 8.1-1(a). The vessel is isothermal
at 25 °C, so the water’s vapor pressure is 3.2 kPa. This vessel has
0.8 l of water with 150 cm<sup>2</sup> of surface area in a total volume of 19.2 l. After 3 min, the air is five percent saturated. What is the mass
transfer coefficient? How long will it take to reach ninety percent
saturation?</p>
<p><a href="https://i.stack.imgur.com/GqX5G.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GqX5G.png" alt=""></a></p>
</blockquote>
<p>The answer starts with this: $$N_1=\frac{\text{Vapor concentration}\cdot\text{Air Volume}}{\text{Liquid Area}\cdot\text{Time}}$$</p>
<p>So far this makes sense. The solution then goes on to do this:</p>
<p>$$N_1=\frac{0.05\cdot(\frac{3.2}{101})\cdot(\frac{1\ \mathrm{mol}}{22.4\ \mathrm{liters}})\cdot(\frac{273}{298})(18.4\ \mathrm{liters})}{(150\ \mathrm{cm^2})(180\ \mathrm{sec})}$$</p>
<p>I understand the denominator but the numerator I'm not sure of. Overall there are 18.4/22.4 mol of gas and 5% of that will be water vapor, I get that. But what's with the temperature and pressure adjustment?</p>
| |thermodynamics|pressure|chemical-engineering|process-engineering|vapor-pressure| | <p>$\frac{1}{22.4}\frac{\mathrm{mol}}{\mathrm{L}}$ is only valid for standard conditions. Use the ideal gas law to calculate the concentration for your conditions.</p>
<p>$s \equiv$ standard conditions</p>
<p>$2 \equiv$ your conditions</p>
<p>$p_s V_s = n_s R T_s$</p>
<p>$\frac{p_s V_s}{n_s T_s} = R = const = \frac{p_2 V_2}{n_2 T_2}$</p>
<p>$\frac{n_2}{V_2} = \frac{p_2 T_s n_s}{p_s T_2 V_s}$</p>
<p>$c = \frac{n_2}{V_2} = \frac{3.2 \mathrm{kPa}}{101 \mathrm{kPa}} \frac{273.15 \mathrm{K}}{298.15 \mathrm{K}} \frac{1 \mathrm{mol}}{22.4 \mathrm{L}}$</p>
<p>Insert in your equation</p>
<p>$N_1 = \dfrac{0.05 \cdot c \cdot V_{Air}}{A \cdot t}$</p>
<p>$N_1 = \dfrac{0.05 \cdot \frac{3.2 \mathrm{kPa}}{101 \mathrm{kPa}} \frac{273.15 \mathrm{K}}{298.15 \mathrm{K}} \frac{1 \mathrm{mol}}{22.4 \mathrm{L}} \cdot 18.4 \mathrm{L}}{150 \mathrm{cm}^2 \cdot 180 \mathrm{s}}$</p>
| 8132 | How do temperature and pressure influence the rate of evaporation of water in a closed vessel? |
2016-03-22T22:08:47.683 | <p>With linear bearings, could they be used to join one rod with another to form a cross-junction and rotate with around a 0.3N-mm torque? Suppose the rods are fairly thin, eg, 5mm in diameter. That is, could linear bearings bear a radial load? Would the bearings deform? </p>
<p><strong>UPDATE</strong> Here's a diagram</p>
<p><a href="https://i.stack.imgur.com/4F5I1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4F5I1.png" alt="enter image description here"></a></p>
<p>Suppose the blue rod is attached to a gear driven by a motor. The blue rod itself is also attached to the bearing's housing. The blue essentially rotates and drives the black, exerting force/stress on the bearing. The idea is to mount a slidable camera on the black which can be rotated by the green gear.</p>
| |mechanical-engineering|bearings| | <p>The bearing should have figures on the moment loads it can take. Failing that, use two bearings and allow them enough movement so effectively each is just supporting a purely radial load.</p>
| 8133 | Could linear bearings bear axial load? |
2016-03-22T23:05:55.630 | <p>In the problem statement below, the question states to find the maximum load $\vec{P}$ that the truss can support. My method of approach was:</p>
<ol>
<li>Draw a FBD for the entire structure.</li>
<li>Generally I would identify all external forces, however for this problem I felt it was unnecessary because I felt they weren't needed if I knew already the forces in each member.</li>
<li>So I then dove straight into joint $D$, and assumed each member connected to $D$ was in tension and experienced the maximum tensile force $\vec{T} = 1500$ lb.</li>
</ol>
<p>However, this led to the wrong answer when solving for $\vec{P}$. The solution manual instead found each external force in terms of $\vec{P}$ and started at joint $A$ to find $\overrightarrow{AD}$ and $\overrightarrow{AB}$ in terms of $\vec{P}$. In addition, in order to get numerical values, the solution manual assumed member $\overrightarrow{AB}$ was experiencing the maximum compression force of 660 lb. However, when I assume that member $\overrightarrow{AD}$ is experiencing the maximum tensile force, it doesn't work out the same.</p>
<p><strong>My question is, conceptually, why must I find each each member's force in terms of $\vec{P}$, and why must one assume $\overrightarrow{AB}$ is experiencing maximum compression (but not $\overrightarrow{AD}$ in maximum tension)?</strong></p>
<p>EDIT: I want to note that I do not need help with solving this problem, nor the math required. I simply am just looking for a conceptual answer as to why my approach did not work (i.e. only analyzing joint $D$).</p>
<p><a href="https://i.stack.imgur.com/7jTDN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7jTDN.png" alt="Problem Statement"></a></p>
| |structures|statics| | <p>The reason is that <strong>you assumed that the elements around node <span class="math-container">$\text{D}$</span> will be the first to fail</strong>. That is not the case. Indeed, it is the elements under compression (<span class="math-container">$\text{AB}$</span> and <span class="math-container">$\text{BC}$</span>) that will fail first. Also, <strong>you assumed that all the members around <span class="math-container">$\text{D}$</span> will present the same axial force</strong>, which is untrue.</p>
<p>To see this, here's your structure with a unitary load (the units are irrelevant):</p>
<p><a href="https://i.stack.imgur.com/oa3Bg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oa3Bg.png" alt="enter image description here"></a></p>
<p>And here are the axial forces in each member under this unitary force (positive in tension):</p>
<p><a href="https://i.stack.imgur.com/NaJ1I.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NaJ1I.png" alt="enter image description here"></a></p>
<p>Since these results are linearly proportional to the force applied, we can now find the concentrated load <span class="math-container">$P$</span> at node <span class="math-container">$\text{D}$</span> that each member can withstand:</p>
<p><span class="math-container">$$\begin{align}
P_{\text{AB}} = P_{\text{BC}} &= \dfrac{660}{0.943} = 700 \\
P_{\text{AD}} = P_{\text{CD}} &= \dfrac{1500}{0.687} = 2183 \\
P_{\text{BD}} &= \dfrac{1500}{1.333} = 1125 \\
\end{align}$$</span></p>
<p>Therefore, the maximum load supported by the structure is the minimum of these values, which is 700 lb.</p>
<p>To show the validity of these results, let's now apply each of these forces and check the results in the relevant members:</p>
<p><a href="https://i.stack.imgur.com/udRpv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/udRpv.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/1PvqI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1PvqI.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/xFQh5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xFQh5.png" alt="enter image description here"></a></p>
<p><sub>All results obtained with <a href="https://www.ftool.com.br/Ftool/" rel="nofollow noreferrer">Ftool</a>, a free 2D frame analysis program.</sub></p>
| 8134 | Where did I go wrong conceptually when attempting to calculate the maximum force on a truss at a given joint? |
2016-03-23T07:54:50.540 | <p>I got a position error measured by a linear encoder like in the picture below.</p>
<p><a href="https://i.stack.imgur.com/wcExZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wcExZ.jpg" alt="Position error graph"></a></p>
<p>Could anyone explain some possible reasons for this kind of shape to occur in control theory?</p>
<p>My system looks like table with four legs.</p>
<p>The four legs move up and down by using motors and have a linear encoder as control reference.</p>
<p>The table should stay in its position after moving up.</p>
<p>The four servo motors actuating the four legs control the position by using linear encoder signals, respectively.</p>
<p>But after moving the table up around 10 mm ~ 20 mm, a strange symptom has happened. One or two encoder signals look like square waves (sometimes one encoder signal, sometimes two, sometimes none).</p>
<p>A simple setup diagram is as below.</p>
<p><a href="https://i.stack.imgur.com/jVGFw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jVGFw.png" alt="Diagram of table"></a></p>
| |control-engineering|control-theory|pid-control| | <p>Your system is not reaching the target value. Possibly this is due to static friction in the actuator mechanism, which causes it not to move until the force exceeds some threshold.</p>
<p>Assuming you are using a PID controller, the residual error will cause the I error term to accumulate slowly. This will cause the actuator force to increase until it is enough to cause actual movement. It seems that this movement will cause a small jerk, which will move the table further than intended and cause opposite error value.</p>
<p>The way you can test this is to change the target height in small steps (e.g. 1 µm) and see what is the smallest distance you can reliably move. You can then either choose to improve the actuator, or to set this as the deadband value in your controller.</p>
| 8139 | Square shaped position error |
2016-03-23T18:48:08.183 | <p>Can a vertical round tube safely support a cantilevered arm mounted near the top of the tube?</p>
<p>ROUND VERTICAL TUBE:</p>
<ul>
<li>6061 Aluminum</li>
<li>2 inch Outside diameter</li>
<li>0.25 inch wall thickness </li>
<li>3 feet height</li>
</ul>
<p>ROUND VERTICAL TUBE / CANTILEVERED ARM INTERFACE:</p>
<ul>
<li><a href="http://dectronusa.com/store/deluxe-2-pole-mount/" rel="noreferrer">this clamp</a></li>
</ul>
<p>(I don’t know the weight of the interface clamp but it is aluminum)</p>
<p>CANTILEVERED ARM:</p>
<ul>
<li>The <a href="http://dectronusa.com/store/rigid-sections/" rel="noreferrer">Arm</a> is 4 feet long with a working load of 100 lbs. at the end of the arm</li>
<li>The arm itself weighs 11.4 lbs</li>
</ul>
<p>ADDITIONAL NOTES:</p>
<ul>
<li><p>The manufacturer of the articulated cantilevered arm (dectron USA) sells a Steel 2 inch OD, 0.125 inch wall thickness tube for this purpose. But their solution won't work for my application.</p></li>
<li><p>I am not worried about the strength of the arm it is rated to support a 100 lb. working load.</p></li>
<li>Don’t think it matters but I have two 12” sections, one 6” and one 18” all rigid</li>
<li><p>The only limit I want is 100 lbs. at the end of the arm other than that I want to use this thing with confidence for decades:)</p></li>
<li><p>Should I consider High-Strength 2024 Aluminum with the same dimensions instead?</p></li>
<li><p>OR could I safely use THINNER walled tubing, 0.125 instead of 0.25 inch wall thickness, 6061 Aluminum, 2 inch OD?</p></li>
</ul>
<p>Thank you very much for looking at this and your help</p>
| |structural-engineering|beam|aluminum|mechanical-failure| | <p>I'm going to run with this assuming the arm looks like the following diagram (I'm ignoring the 11.4 pounds of the bar for now to make the concepts easier to explain - that can be added later by assuming all 11.4 pounds run through the center of the bar.):</p>
<p><a href="https://i.stack.imgur.com/xLAkg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xLAkg.png" alt="enter image description here"></a></p>
<p>This is a pretty simple setup. To figure out the loading on the shaft, we have to move the load from the tip of the arm to the center of the shaft. Fortunately, we can use the principle of <a href="https://en.wikipedia.org/wiki/Couple_(mechanics)" rel="nofollow noreferrer">force couples</a>. The basis is that a force off-center can be moved to the center, but you have to add a couple, or moment, equal to the force times the distance moved. Here is what I came up with.</p>
<p><a href="https://i.stack.imgur.com/UjVz7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UjVz7.png" alt="enter image description here"></a></p>
<p>Now we can apply some basic formulas. We need to look at the cross section of the hollow tube and get some <a href="http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesHollowCircle.aspx" rel="nofollow noreferrer">basic properties</a> of the cross section, as well as on the <a href="http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA6061t6" rel="nofollow noreferrer">material</a> (we need yield strength $\sigma_y$, and $E$, the young's modulus). The important sectional properties are the area, $A$, the area moment of inertia, $I_{xx}$, and the Section Modulus, $S_{xx}$. Formulas for these can be described below (site also has a <a href="http://www.amesweb.info/SectionalPropertiesTabs/SectionalPropertiesHollowCircle.aspx" rel="nofollow noreferrer">calculator</a>).</p>
<p><a href="https://i.stack.imgur.com/wP99P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wP99P.png" alt="enter image description here"></a></p>
<p>Now we can start looking at the four methods of failure:</p>
<ul>
<li>Critical Vibration <a href="https://en.wikipedia.org/wiki/Resonance" rel="nofollow noreferrer">(Natural Frequency)</a></li>
<li>Deflection</li>
<li>Stress</li>
<li>Buckling</li>
</ul>
<p>We can eliminate <strong>vibration</strong> immediately as a method of failure, but for reference, the critical frequency for a cantilevered beam with an end weight only is: (Ref <a href="http://rads.stackoverflow.com/amzn/click/0071742476" rel="nofollow noreferrer">Roark's Formulas for Stresses and Strains, Eighth Ed.</a>, Table 16.8, Case 3a) $$f = \frac{1.732}{2\pi}\sqrt{\frac{EI_{xx}g}{WL^3}}$$
$$f = {0.2757}\sqrt{\frac{10000000 \frac{lbf}{in^2}*0.537in^4*386\frac{in}{s^2}}{100lbf*(36in)^3}} = 5.811 Hz$$</p>
<p><strong>Deflection</strong> in z (down the pole) is minimal. In this case then: $$\delta_z = \frac{W*L}{E*A}$$ $\delta_z = \frac{100lbf*36in}{10000000 \frac{lbf}{in^2}*1.374{in^2}} = 0.000262 in$ which is clearly negligible. <a href="https://en.wikipedia.org/wiki/Deflection_(engineering)" rel="nofollow noreferrer">Deflection</a> in x/y (across the body) is likely not. In this case $$\delta_x = \frac{ML^2}{2EI_{xx}}$$ (Ref Roark again, Table 8.1 Case 3a). $$\delta_x = \frac{5000 in*lbf * (36in)^2}{2*10000000\frac{lbf}{in^2}*0.537in^4} = 0.603 in$$ This is fairly significant and should be considered (the column will tip over just shy of 5/8").</p>
<p><strong>Stress</strong> is also a concern here. We need to look at the stress from the bending moment as well as from the downward force. In this case, the formula is: $$\sigma = {F \over A} \pm \frac {M}{S_{xx}}$$ (Ref: <a href="https://en.wikipedia.org/wiki/Bending#Large_bending_deformation" rel="nofollow noreferrer">Wikipedia</a>).
$$\sigma = \frac{100 lbf}{1.374 in^2} \pm \frac{5000 lbf*in}{0.537in^3} = 72.78psi \pm 9310.98 psi$$</p>
<p>Two things to note - the bending part dominates over the downward portion, so this is the most important part to look at in a preliminary analysis. The second is that this is pretty high, but not terrible. It has a 4.26 F.O.S, which is better than 1.8 typically used in most structures - but aluminum has a <a href="https://en.wikipedia.org/wiki/Fatigue_limit" rel="nofollow noreferrer">fatigue life</a> that needs to be investigated. Repeated quick loading (pushing on it and pulling within seconds) could cause break, but it should have a decent life. This, however, is a separate question entirely.</p>
<p><strong><a href="https://en.wikipedia.org/wiki/Buckling" rel="nofollow noreferrer">Buckling</a></strong> is where things get interesting - we need to account for both the moment buckling factor of safety and the downward force factor of safety separately, and combine. Read the Wikipedia article for more information about buckling, but suffice it to say that, with the exception of metal bars (like you have here), it's mostly empirical and relies on a lot of testing. I'll put the formulas here to cover the details so you can get going on further testing if needed. The formula for the downward force is: $$P' = 0.25*\frac{\pi^2*E*I}{L^2}$$ Source: Roark's again, Table 15.1, Case 1a. This leads to $P' = 40895 lbf$, or a FOS of $\frac{40895lbf}{100lbf} = 409$. Bending is $$M' = 0.72*\frac{Ert^2}{\sqrt{1-\nu}}$$ (Source is Roark's again, Table 15.2, Case 16) $\nu$ for aluminum is <a href="http://www.engineeringtoolbox.com/poissons-ratio-d_1224.html" rel="nofollow noreferrer">0.35</a>. This leads to $M' = 558312 in*lbf$, or FOS of $\frac{558312 in*lbf}{5000 in*lbf} = 111.6$. Combining the two is $$\frac{1}{\frac{1}{111.6} + \frac{1}{409}} = 87.7$$ Therefore, buckling is not a method of failure. (Anything over 5 is usually acceptable without further analysis). <strong>Therefore, the 2" aluminum tube is satisfactory for the given loads</strong></p>
<p>Now onto the additional notes:</p>
<p>Should I consider High-Strength 2024 Aluminum with the same dimensions instead? <strong>You should be fine with this design</strong></p>
<p>Could I safely use THINNER walled tubing, 0.125 instead of 0.25 inch wall thickness, 6061 Aluminum, 2 inch OD? <strong>This I leave as an exercise for you. It should be pretty easy to investigate with the tools given - though note the deflection was already fairly high.</strong></p>
| 8144 | Strength of a Vertical Round Aluminum Tube with load cantilevered off it? |
2016-03-24T07:50:00.190 | <p>What are the criteria for choosing repeating variables in Buckingham's Pi theorem in dimensional analysis? In many problems, it's solved by taking D,V,H (Diameter, Velocity, Height) as repeating variables. Why do they take the above variables as repeating variables, when the problem also contains the following variables g,u (acceleration due to gravity, viscosity)?</p>
| |fluid-mechanics| | <p>The repeating variables are any set of variables which, by themselves, cannot form a dimensionless group. Diameter, velocity, and height cannot be arranged in any way such that their dimensions would cancel, so they form a set of repeating variables.</p>
<p>For example, let's assume that we suspect that a fluid we're studying behaves as a function of several variables including a characteristic length, velocity, viscosity, density, surface tension, etc., and we want to see what dimensionless numbers we can make out of them. A possible choice of repeating variables would be length ($l$), velocity ($v$), and density ($\rho$) (in MKS units they would be $m$, $\frac{m}{s}$, $\frac{kg}{m^3}$), because they cannot be combined in any way to make a dimensionless group. Now add surface tension ($\sigma$, units of $\frac{kg}{s^2}$) and combine all four variables to make a dimensionless group like this:</p>
<p>$$
\frac{\rho l v^2}{\sigma}
$$</p>
<p>This is the <em>Weber Number</em>. Another possibility would be to use viscosity instead of surface tension:</p>
<p>$$
\frac{\rho l v}{\mu}
$$</p>
<p>Which is of course the <em>Reynolds Number</em>. Ultimately, the choice of which combination to use out of all the possibilities comes down to whichever is more useful for the type of problem you're working on; Reynolds is good for turbulence and heat transfer, while Weber is more suitable for bubble and droplet formation.</p>
| 8146 | Choosing of repeating variables in Buckingham's Pi theorem |
2016-03-24T12:54:47.907 | <p>I think there is an error in this questions solution and I want someone to check:</p>
<blockquote>
<p>Example 8.1-4:Mass transfer from an oxygen bubble A bubble of oxygen originally 0.1 cm
in diameter is injected into excess stirred water, as shown schematically in Fig. 8.1-1(d).
After 7 min, the bubble is 0.054 cm in diameter. What is the mass transfer coefficient?</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/blLBF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/blLBF.png" alt="enter image description here"></a></p>
<p>The solution goes on to state:</p>
<p>$$\frac{d}{dt}(c_1\frac{4}{3}\pi r^3)=AN_1=-4\pi r^2k[c_1(\text{sat})-0]$$
$$\frac{dr}{dt}=-k\frac{c_1(\text{sat})}{c_1}=-0.034k$$</p>
<p>Shouldn't it be:
$$\frac{dr}{dt}=-3k\frac{c_1(\text{sat})}{c_1}=-0.102k$$</p>
| |thermodynamics|chemical-engineering|process-engineering|vapor-pressure| | <p>Your approach is wrong because the left hand side represents the derivative of the composition of two time dependent functions. You completely ignored that fact and started cancelling terms. I try to give a simpler example:</p>
<p>$\frac{d 4x^3}{dx} = kx^2
\Leftrightarrow 12x^2=kx^2
\Leftrightarrow 12=k$</p>
<p>This is not the same as</p>
<p>$\frac{d 4x}{dx} = k \Leftrightarrow 4=k$</p>
<p>Therefore use the chain rule</p>
<p>$\frac{d(c \cdot V)}{dt} = V\frac{dc}{dt}+\frac{dV}{dt}c=-Ak\Delta c$</p>
<p>I would now assume a stationary diffusion, hence</p>
<p>$V\frac{dc}{dt} = 0$</p>
<p>$\frac{dV}{dt}c=-Ak\Delta c$</p>
<p>$A = \frac{dV}{dr}$</p>
<p>$\frac{dV}{dt}c=-\frac{dV}{dr}k\Delta c$</p>
<p>For this step I hope there are no mathematicians around, I cancel the differentials</p>
<p>$\frac{dr}{dt}c=-k\Delta c$</p>
<p>$\frac{dr}{dt}=-k\frac{\Delta c}{c}$</p>
| 8148 | Mass Transfer from an Oxygen Bubble |
2016-03-24T13:53:53.253 | <blockquote>
<p>A steam power plant employs two adiabatic turbines in series. Steam
enters the first turbine at 650 C and 7000 kPa and discharges from the
second turbine at 20 kPa. The system is designed for equal power
outputs from the two turbines, based on a turbine efficienct of 78%
for each turbine. Determine the temperature and pressure of the system
in its intermediate state between the two turbines. What is the
overall efficiency of the two turbines together with respect to
isentropic expansion of the steam from the initial to final state.</p>
</blockquote>
<p>How should I determine the pressure between two turbines?I have the <a href="https://drive.google.com/file/d/0B2xNtfJLGsLwZ3ZnNWxQS283ZXM/edit" rel="nofollow">solution</a> from a solution manual (prob. 8.6) but I just can't understand why it takes pressures around 700 KPa and does an interpolation.
what we know:</p>
<p>$S3=S2=S1$ (if working isentropically) ($S$ is entropy)</p>
<p>$x$ (Quality of exiting vapor)</p>
<p>Note:we could determine properties of the intermediate flow properties using one property($S$) just if it's two-phase, but it's assumed that the intermediate flow is superheated.</p>
<p>I'm sure of problem's data and that's all we have.</p>
| |mechanical-engineering|thermodynamics| | <p>The answer in the solution manual simply assumes a set of four values of intermediate pressure (that somehow magically the actual intermediate pressure lies within!):
$$P_2 = \begin{pmatrix}725\\750\\775\\800\end{pmatrix} \text{kPa}$$</p>
<p>And for each pressure in the set, the one that results in an equal power output from the two turbines (or the closest value to $\triangle W = 0$) is the actual intermediate pressure (After some trials).</p>
<blockquote>
<p>and we interpolate linearly to find the value of $P_2$ for which the work difference is zero:</p>
</blockquote>
| 8149 | Determining intermediate pressure between two turbines |
2016-03-26T17:10:25.213 | <p>Suppose the fit for a rod and a hole in a gear is supposed to be free running and the fit is 3mm H9/d9. For the specification of the hole for the manufacturing of the gear, is "3mm H9" already adequate since d9 only refers to the shaft tolerance which may not be pertinent to the manufacturing of the gear? Sometimes I see the fit specified on drawing sheets for some reasons, e.g., "10mm H9/d9".</p>
<p>If my understanding is correct, 3mm H9 essentially means 3mm - 3.025mm and 3mm d9 means min. 2.955mm - 2.98mm. So the clearance ranges from 0.02mm to 0.07mm - am I right?</p>
| |design|tolerance| | <p>Your clearance range is the minimum hole diameter minus max shaft diameter to max hole diameter minus minimum shaft diameter. </p>
<p>Answer: Yes. Your math seems correct. </p>
<p>This is a useful tool: <a href="http://www.amesweb.info/FitTolerance/FitTolerance.aspx" rel="nofollow">http://www.amesweb.info/FitTolerance/FitTolerance.aspx</a></p>
| 8173 | Specification of hole design tolerance |
2016-03-27T17:15:19.463 | <p>I am working on a project and I have some questions about setting up a passive heating system (thermosiphon) with a heat exchanger.
I will have a main boiler with a coil of copper tubing inside. This coil will be my heat exchanger. It will then continue to pass through an object which is the target of the passive heating cycle. My question is about the placement location of the coil's ends.
Please see the attached system drawings. Will either of these work? Also, to clear up the "T" in the lower-left: for the purposes of discussing the thermosiphon loop, you can pretty much ignore this. Check valves will be in place to ensure the return goes back into the exchanger coil.
Thanks in advance for the help<img src="https://i.stack.imgur.com/dMULW.jpg" alt="![enter image description here">]<a href="https://i.stack.imgur.com/dMULW.jpg" rel="nofollow noreferrer">1</a></p>
| |heat-transfer|heat-exchanger| | <p>In the broad sense your question is asked I'd say yes, this will work either way. </p>
<p>Two things to consider that could be an indicator for you to go with #1 or #2:</p>
<p>Depending on the actual design, the isolation of the copper coil in between the heater and the item to be heated you might try and keep the distance as short as possible to reduce heat loss. So this might favor #2</p>
<p>On the other hand #1 may be favorable if you need to increase heat transfer in the tank and want to increase the surface area for heat transfer.</p>
| 8179 | Thermosiphon System |
2016-03-28T12:33:59.117 | <p>Consider a fixed steam engine, such as in a Lancashire cotton mill or pumping in a Cornish mine. Then consider a fixed diesel engine such as a large generator or ship's engine. </p>
<p>Why does the steam engine have one or two large cylinders with long stokes, but the diesel has many cylinders with relatively short strokes?</p>
| |steam|diesel| | <p>The difference of few large cylinders or many small cylinders has less to do with running on steam or diesel, but more with the era in which the engines were built. Early engines (both steam and diesel) had few, large cylinders, probably because that is how engineers tried to make more powerful engines: make everything bigger. The realization that multiple smaller cylinders gives a smoother running engine and more power output (because small pistons can run at higher rpm) came later, and by that time steam cylinder engines were being replaced with either diesels or steam turbines. But smaller cylinder <a href="https://en.wikipedia.org/wiki/Steam_motor" rel="nofollow noreferrer">steam motors</a> also existed. And if you search for videos of people running antique diesel engines, those are also often single or few cylindered.</p>
<p>Length of the stroke is not really relevant, what matters is the ratio between the minimum and maximum volume, and you can increase that by having a long stroke with a large maximum volume, or a very small minimum volume. Modern engines use short strokes because those allow for higher rpms. </p>
<p><em>addition:</em> Another factor is that a single cylinder engine is easier to build than a multi cylinder engine. In the 19th century the technology for building engines (both steam or internal combustion) was less advanced, and the simpler designs were preferred. As tools and construction methods improved, and as more uses for light high speed engines became apparent, engine designs moved from single large cylinders to multiple smaller cylinders. This too has little to do with the engine type, but much more with the era in which the engine was built.</p>
| 8193 | Why do steam and diesel engines have differing cylinder configurations? |
2016-03-28T13:30:42.413 | <p>I'm using AutoCAD 2015, and I run into this all the time. I'm given a group of DWG drawings, and I want to plot them into one monochrome PDF, but they do not have monochrome.ctb selected as the plot style. Is there any way in Batch Plot to publish to PDF using monochrome without editing the individual DWG files? If not, is there a way to edit the page setup of all the drawings together as a group?</p>
| |autocad| | <p>After playing around with it for a while, this seems to work:</p>
<ol>
<li>In the Batch Plot dialog, add all the drawings/sheets you want to include:</li>
</ol>
<p><a href="https://i.stack.imgur.com/y57FU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y57FU.png" alt="Batch Plot dialog"></a></p>
<p>In the picture above, you can see that different drawings have different default Page Setups. None of these drawings have a page setup I can use.</p>
<ol start="2">
<li>Select the first drawing, and under Page Setup, select "Import..."</li>
</ol>
<p><a href="https://i.stack.imgur.com/cbEuH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cbEuH.png" alt="Select "Import...""></a></p>
<ol start="3">
<li><p>Choose a DWG file that does contain a page setup you like. In my case, I'm looking for one that will plot the Extents of the model using the monochrome.ctb plot style.</p></li>
<li><p>Your imported Page Setups are now available in the list. Select the one you want. In my case, I'm choosing the one called "Imported: 11x17".</p></li>
</ol>
<p><a href="https://i.stack.imgur.com/WYydj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WYydj.png" alt="Select the Page Setup"></a></p>
<ol start="5">
<li><p>Select this Page Setup for each drawing. You can select all the sheets at once (Ctrl-A, or Shift-Click) and then choose a page setup.</p></li>
<li><p>Save the Sheet List. This will preserve your imported Page Setup for the entire list, so you won't need to do these steps in the future for these drawings.</p></li>
<li><p>Click the Publish button to plot the drawings.</p></li>
</ol>
<p>This procedure doesn't make any changes to the DWG files. In the future, you can load the Sheet List, and the imported Page Setups will still be selected.</p>
| 8195 | How do I batch plot drawings in monochrome without editing DWG? |
2016-03-28T13:30:59.527 | <p>I am currently working on a project (or actually returning to a project) and need to have a heat exchanger boiler made. The hx boiler will be relatively small, a 10" long cylinder around 4" in diameter. One end can be capped, but the other should have a flange so I can work inside the unit. Inside, I will have a coil of copper tubing (heat exchange) and various other sensors/ports. I don't have access to a full machine shop to fabricate such a boiler myself so I am contacting a number of companies. My question really is in regards to the thickness of the steel required for the boiler. The boiler, at max pressure, will only be at 2.5 bar. Additionally, the material must be food-safe; the main boiler will contain water that will be used in food preparation. Given these working parameters, of what type and how thick should the stainless steel be? I will attach various rough sketches of the system for extra information.
Thanks in advance for the help!
<a href="https://i.stack.imgur.com/g33zq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g33zq.jpg" alt="Flange Cap"></a>
<a href="https://i.stack.imgur.com/EnJ9c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EnJ9c.jpg" alt="Rough Cutaway"></a></p>
| |materials|boilers| | <p>In addition to the excellent answers on pressure vessels already posed a steam boiler has the additional complication that, as you are dealing with a change in phase under pressure there is a particular hazard with steam that you can get very rapid pressure spikes under certain conditions so the safety regulations for steam boilers are rather more stringent than for pressure vessels in general. For example interruption of the flow of water may cause a rapid pressure rise which may not be immediately apparent see <a href="https://en.wikipedia.org/wiki/Steam_explosion" rel="nofollow">steam explosion</a> </p>
<p>You will also find that your jurisdiction requires specific welding codes for welding of steam pressure vessels and/or individual testing and inspection. </p>
| 8196 | Questions about making stainless steel boiler |
2016-03-29T06:12:02.857 | <p>I am designing a wind turbine to generate 1300 W.</p>
<p>I modeled the output voltage of an AC generator using the following specifics:</p>
<ul>
<li>Number coil turns = 300.</li>
<li>Copper wire gauge = 1.02 mm.</li>
<li>Knowing the density of copper, I calculated the weight of the coil to be 4.39 kg.</li>
<li>Magnetic field strength = 1.24 T. (<a href="http://www.arnoldmagnetics.com/Portals/0/Files/Catalogs%20and%20Lit/Neo/151021/N38AH%20-%20151021.pdf?ver=2015-12-07-103811-373" rel="nofollow">here are the specs for the magnet</a>)</li>
<li>Area = 1 m<sup>2</sup>.</li>
<li>Rotational speed = 45 RPM.</li>
</ul>
<p>I plugged these numbers into the output voltage equation:</p>
<p>$$\epsilon = NBA\omega\sin(\omega t)$$</p>
<p>to get a value of 1753 V. I am assuming that my generator will generate at least the 0.74 A to output a power of 1300 W.</p>
<p>Then, I modeled my turbine. I used the equation:</p>
<p>$$P_T = \frac{1}{2}\rho A v^3 C_p$$</p>
<p>to calculate rotor power. I assumed:</p>
<ul>
<li>My turbine would be 30% efficient.</li>
<li>The wind velocity would be 10 m/s.</li>
<li>The area swept by blades would be 2.63 m<sup>2</sup>.</li>
<li>The density of air would be 1.25.</li>
</ul>
<p>After plugging these values in, I get a power yield of 493 W. However, I find that the rotor speed is well above 45 RPM. This equation gives the tip speed ratio of the turbine:</p>
<p>$$\lambda = \frac{wr}{v}$$</p>
<p>Knowing that the optimum tip speed ratio of a three-blade turbine is 4.2, radius of the rotor blade is 0.915 m, and velocity of air is 10 m/s, that angular velocity would be 45.9 rad/s or 440 RPM.</p>
<p>How can I get 1300 W from this turbine when the theoretical yield from power of the rotor is only 493 W, even though my rotor should turn at the correct RPM for the generator to produce 1300 W?</p>
| |mechanical-engineering|turbines|wind-power| | <p>I don't think you really designed a 1300 W wind turbine. You <em>assumed</em> your generator would be able to produce the necessary current but to be able to produce 1300 W you have to build a wind turbine big enough, given its efficiency. </p>
<p>At $C_p=0.3$ and $v=10\ \mathrm{m/s}$, you simply need an area of:
$$A=\frac{2P_T}{C_pv^3\rho}=7.07\ \mathrm{m^2}$$
$$\Rightarrow R=1.5\ \mathrm{m}$$</p>
<p>Moreover, you say $\lambda=4.2$ is optimal for a three-blade turbine, but it actually depends on the blade profile, which is optimized for a given "design tip speed ratio". For more information about this, see chapter 3.7 of "Burton, <em>Wind energy handbook</em>, 2011". </p>
<p>See also "Duquette, <em>Numerical Implications of Solidity and Blade Number on Rotor Performance of Horizontal-Axis Wind Turbines</em>, 2003" about influence of blade number on power coefficient.</p>
| 8207 | Why does my model of a wind turbine show high RPM with low power output? |
2016-03-30T12:06:20.637 | <p>I am making a testing facility for pumps. This facility has to be able to handle 500 m<sup>3</sup>/h with at little turbulence as possible. The basin itself is 5 meters long, 2,5 meters wide and 2 meters tall. The water level in the basin is kept at 1,9 meters.</p>
<p>The basin is designed in two sections: the first is where the discharge flow comes into the basin (pipe ends about 1 meters in from the bottom). In that area the water flows freely.</p>
<p>After that it has to flow over the retaining wall (1.7m high), which should create a somewhat uniform flow.</p>
<p><a href="https://i.stack.imgur.com/5IV2J.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5IV2J.png" alt="overview"></a></p>
<p>Are there better solutions to do this? </p>
<p>One other idea I have is a retaining wall somewhat like this:</p>
<p><a href="https://i.stack.imgur.com/0tz0T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0tz0T.png" alt="wall idea"></a></p>
<p>This makes for three plates with cutout stripes.</p>
<p>The idea behind this is to have a more uniform flow across the entire basin instead of a overflow, making it create a small bit of turbulence there.</p>
| |water-resources|turbulence|flow-control| | <p>Assuming you are wanting to minimize turbulence in the region of the pump, you would probably need to use flow straighteners somewhere upstream of the pump. You would have to study the optimal diameter and length of the straighteners. MIT simply used <a href="https://en.wikipedia.org/wiki/Flow_straightener#A_low-cost_handmade_flow_straightener" rel="nofollow">drinking straws</a> to do this in a wind tunnel. Here are a couple of the more well known papers on the subject:</p>
<ul>
<li><a href="http://fluidsengineering.asmedigitalcollection.asme.org/article.aspx?articleid=1433483" rel="nofollow">Reducing Water Tunnel Turbulence by Means of a Honeycomb</a> by Lumley and McMahon</li>
<li><a href="http://fluidsengineering.asmedigitalcollection.asme.org/article.aspx?articleid=1423349" rel="nofollow">Control of Free-Stream Turbulence by Means of Honeycombs: A Balance Between Suppression and Generation</a> by Loehrke and Nagib</li>
</ul>
| 8221 | Best way to get rid of turbulent flows? |
2016-03-30T13:44:17.563 | <p>I have the situation from the following drawing:</p>
<p><a href="https://i.stack.imgur.com/gHSf2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gHSf2.png" alt="enter image description here"></a></p>
<p>It represents a model for measuring the torque of a generator.
There is an arm connected to stator and, the other end of the arm to a bending type load cell. I want to model this system and have as an input the torque of the generator rotor and as output the torque that reaches the load cell. I want to model somehow how is the transfer of energy from the rotor to the load cell.
The load cell type is Hottinger Z6FC3.
The system is connected like in the figure below.</p>
<p><a href="https://i.stack.imgur.com/q7XuX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q7XuX.png" alt="enter image description here"></a></p>
<p>I am new at this and don't know much mechanics. I don't know how to write the motion equations for the two bodies and connect them.</p>
| |mechanical-engineering|torque| | <p>You need to apply <a href="https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion" rel="nofollow">Newton's second law</a> that states:</p>
<blockquote>
<p><strong>Second law</strong>: The vector sum of the forces F on an object is equal to
the mass m of that object multiplied by the acceleration vector a of
the object: F = ma.</p>
</blockquote>
<p>In your case, you need to replace force by torque, mass by inertia and acceleration by angular acceleration. Given that the torque provided by the rotational springs is the spring stiffness x the relative angle and that the torque provided by the rotational dampers is (in the first instance) the damping coefficient x the relative angular velocity, you get:</p>
<p>$ T = (J_s + J_a) \ddot{\theta_r} + C_m (\dot{\theta_r} - \dot{\theta_s}) + K_m (\theta_r - \theta_s)$</p>
<p>$ J_{Gs} \ddot{\theta_s} + C_{Gs} \dot{\theta_s} + K_{Gs} \theta_s = C_m (\dot{\theta_r} - \dot{\theta_s}) + K_m (\theta_r - \theta_s)$</p>
<p>where $\theta_r$ and $\theta_s$ are the rotational displacements of the rotor and the stator, respectively.</p>
| 8222 | motion equation of rotor and stator |
2016-03-30T16:35:47.030 | <p>We have obtained an old mercury-vapor <a href="https://en.wikipedia.org/wiki/Diffusion_pump">diffusion pump</a> for high vacuum. However, we do not want to operate it with mercury in our lab due to health concerns (which is presumably the reason it was scrapped by the original owner). </p>
<p>Instead, we plan to fill it with a silicone oil of the <em>DC-704</em> type or similar. The pump has thermal regulation and a large baffle to stop oil vapor counter-propagation.</p>
<p>Has anybody tried replacing mercury with oil? What are possible problems with this approach? </p>
| |pumps|safety|vacuum|vacuum-pumps| | <p>I did some internet searching and did not find anyone that had done a direct swap out. Granted, probably the majority of these replacements occurred before the existence of the internet ;-). As I am sure you are already aware (but to be thorough), modern diffusion pump designs do use <a href="http://www.vpcinc.ca/Products/Lubricants/Diffusion_Fluids.html">synthetic working fluids</a>. </p>
<p>It will probably take some fine tuning of the temperatures of the heater and cooler, but I don't see any reason why it wouldn't work. If you have means of measuring the vacuum level, you could run through some different temperatures to empirically find its new ideal operating conditions.</p>
<p>Worst case, the pumping efficiency will be less because the nozzle diameter was designed for higher density particles. I am not a vacuum physicist however, so who knows, it may work even better than before ;-)</p>
<p>Another possible issue is that it may have more <a href="http://www.mtm-inc.com/av-20121127-diffusion-pumps.html">backstreaming</a> of the oil into the vacuum chamber resulting in unacceptable levels of oil contamination. Maybe the baffle you mentioned will take care of this; hard to say without some experimentation.</p>
<p>Just as long as no oxygen reaches the oil during operation, it wont fail catastrophically, so give it a try and let us know how it went!</p>
<p>Reference:<br>
<a href="http://www.xtronix.ch/PDF/Diffusion%20Pump.pdf">Good overview of diffusion pump working fluids</a><br>
<a href="http://vacaero.com/information-resources/vacuum-pump-practice-with-howard-tring/1403-oil-diffusion-pump-controls.html">Some good history or mercury as a diffusion pump working fluid</a><br></p>
| 8224 | Replace mercury with silicone oil in diffusion pump |
2016-03-31T07:55:06.040 | <p><strong>Important note</strong>: I am writing a structural FEM software from scratch, so I won't able to use existing commercial FEM package. I would need to know how all the important quantities are calculated from the first principles.</p>
<p>I would need to determine the <strong>accidental torsional effect</strong> ( <a href="https://drive.google.com/file/d/0B5oarfYUwEDrY1NBMU52M3FHVDA/view" rel="nofollow noreferrer">UBC97, 1631.5.6, pp-27</a>, <strong><em>this accidental torsion effect is needed for both response spectrum and linear time history analysis</em></strong>) for seismic dynamic analysis.</p>
<p>$M_t=F_xe_y$</p>
<p>Where </p>
<p>$F_x$ is the equivalent lateral static force in $x$ direction</p>
<p>$e_y$ is the accidental eccentricities in $y$ direction</p>
<p>Take note that all these involved quantities are evaluated <strong><em>on floor basis</em></strong>. </p>
<p>My structures are quite irregular and general, such as this:</p>
<p><img src="https://i.stack.imgur.com/z6Mu3.jpg" alt="enter image description here"></p>
<p>For a cantilever column or an one dimensional lumped mass model, the term <strong>on floor basis</strong> is easy to interpret because there is only 1 node per floor, and the $F_x$ can only be applied on this node itself. Compare this lumped mass model with the above 2D structure FEM model and you can see the difference:</p>
<p><a href="https://i.stack.imgur.com/SauGW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SauGW.jpg" alt="enter image description here"></a></p>
<p>But when you have such a 2 dimensional irregular structure, analyzed using FEM, calculating $F_x$ and $M_t$ is no longer easy:</p>
<ol>
<li>How should we obtain $F_x$, given that it is lateral <em>static</em> force, but we are doing dynamic analysis?</li>
<li>There are so many nodes on a floor, so on which node(s) should the $M_t$ be applied? </li>
<li>In FEM, each beam, column, wall and slab are multiple-node elements. And each node has a internal force value resulted from FEM analysis, what should I do with all these forces at different nodes? <a href="https://engineering.stackexchange.com/questions/7662/how-to-derive-equivalent-static-load-for-irregular-structure-modeled-with-fem#comment14210_7673">The comment here</a> says that I should sum up all of the forces on each related node in order to obtain the correct $F_x$ ( I'm not sure whether I interpret it correctly or not). But I am highly skeptical of this approach, because won't that make $F_x$ dependent on the number of nodes, an artifact of how we mesh an element? How can a real physical quantity increases with the number of nodes we decide to mesh?</li>
<li>If 3) is true, then what about the $F_x$ on the wall above a particular floor? Should we sum them in? What about the internal slab and beam? Do they have this force also?</li>
<li>How should we intepret $M_t$? Is it internal member force to be resisted by beam/slab? If yes, if there are many slabs and beams into $x$ direction, then how should this $M_t$ be distributed? Or is it an external static force; so one would have to apply this external force on the structure in order to calculate the individual internal member force induced by it ( via the FEM equation $ku=F$)? </li>
</ol>
<p>So the question is, how to calculate and interpret accidental torsion in a FEM model? I've tried to obtain a general way to reduce a 2 Dimensional general FEM structure into 1 Dimensional lumped mass model, so that I can apply the above formulation easily, but <a href="https://engineering.stackexchange.com/a/7673/3353">I was told that this wasn't possible</a>. </p>
| |structural-engineering|structural-analysis|seismic| | <p>There is a paper on how to account for <a href="https://wiki.csiamerica.com/download/attachments/9536133/Fahjan%2C%20Accidental%20eccentricity.pdf?version=1&modificationDate=1338950028768&api=v2" rel="nofollow">accidental eccentricity in dynamic modal analysis of buildings</a>:</p>
<blockquote>
<p>The methodology is based on dynamic modal superposition technique by
applying of accidental torsion to global force vectors related to each
modal shape. For this purpose, the global displacement vectors are
computed for each modal shape at the first step. Then, corresponding
global force vectors are computed by using the global mass matrix, the
eigenvalues of the system and the global displacement vectors. For
each mode shape, the accidental torsional moments of the global force
vector are updated by the required amount of eccentricity in either
direction. Static analysis is carried out to find the modified global
modal displacement vectors and the internal forces for each member for
each modal shape. The nodal displacements and the internal force
resultants can be combined by using standard modal combination
techniques.</p>
</blockquote>
| 8234 | Accidental Torsion and Equivalent Static Load for an irregular structure |
2016-04-01T00:56:08.217 | <p>I have a 10 mm diameter rod of 99.5% pure tungsten that I want to drill a 0.1 mm hole in using my Grizzly G0768 lathe. The hole only needs to be between 0.1 mm and 1 mm deep and is for a 3D printer hot nozzle. I don't know if this will break the bit yet, but I will try it when the drill chuck comes in the mail. If the bit breaks, I'll move up to a 0.15 mm hole, and so on...</p>
<p>I can't afford a high precision mill that costs more than $1,000. Assuming my lathe can't drill such a small hole, what other options do I have to accomplish this task on a budget? When I have gone shopping I have seen no indication of the stability / precision of the tools.</p>
<p>I could send the pieces of tungsten off to get drilled, but I expect that might cost a fortune.</p>
<p>If anyone has experience making holes this tiny, I would like to know the approach you used.</p>
<p>I have nice $12, 0.1 mm PCB bits made in Germany from Tungsten Carbide that break if you so much as look at them. I also found some cheaper bits on eBay but they don't look like drill bits so I'm not sure if they'd even cut a hole.</p>
| |machining|drilling|lathe| | <p>Call local machine shops, and ask them, if they have a sonic mill. It uses special bits with coolant and ocelats to make the cuts. It can cut accurately through 1/2 inch in about 45 minutes. It shouldn't be too expensive, basically it vibrates the the material.</p>
| 8246 | How can I cheaply drill a 0.1 mm hole in a small piece of tungsten? |
2016-04-01T09:14:06.017 | <p>I am working on a project to bring light into a bunker. The bunker is 3-4 meters below the ground.</p>
<p>I am looking into bringing natural light into the space, instead of using electricity. The space is difficult to access and light must therefore be transported by flexible cable for about 20 meters from the surface to the areas I want to light up. I am thinking of pipe around 1 cm diameter maximum.</p>
<p>I am looking into a cheap solution to create an optical pipe that is effective at transporting light.</p>
<ol>
<li>I am wondering what techniques are typically used.</li>
<li>What kind of pipes would you use ?</li>
<li>Is it possible to fill the pipe with other materials to increase its effectiveness ?</li>
</ol>
| |fiber-optics| | <p>You can send your light via internal total reflection. You need to research a flexible material that's hollow inside (to fill with liquid) and the material's index of refraction match the formula for total internal reflection.</p>
| 8250 | Create a DIY light tube |
2016-04-02T18:34:07.813 | <p>Here is some background to the problem (in a stirred tank):</p>
<p>"With yield stress non-Newtonian (viscoplastic) fluids, it is possible to generate an agitated volume around the impeller, defined as a cavern, surrounded by a stagnant region where the shear stress is insufficient to overcome the apparent yield stress of the fluid."</p>
<p>Sometimes you can get a cylindrical cavern around the impeller, see the below image. <a href="https://i.stack.imgur.com/YeL2m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YeL2m.png" alt="Cylindrical Cavern around impeller"></a></p>
<p>"By performing a force balance between the applied torque, Γ and the shear stress acting on the surface of a cylinder, we can define the boundary by setting the shear stress equal to the yield stress τ = τY. The total torque is given by:
$$\Gamma = \frac{\pi}{2} \tau_{y}H_{C}D_{C}^2+\frac{\pi}{6}\tau_{y}D_{C}^3$$</p>
<p>I just can't get the second term. The first term I can get by doing:
$$\Gamma_{1}=\tau_y \cdot Area_{Curved} \cdot \frac{D}{2} = \pi \cdot \frac{D^2}{2} \cdot H_{c} \cdot \tau_{y}$$</p>
<p>This gets me the first term...but the second term I just can't get, this is what I'm doing:</p>
<p>$$\Gamma_{2}=\tau_{y} \cdot Area_{Faces} \cdot \frac{D}{2} =\tau_{y} \cdot 2 \pi \cdot \frac{D^2}{4} \cdot \frac{D}{2} = \tau_{y} \cdot \pi \cdot \frac{D^3}{4} $$</p>
<p>Argh, so I'm getting D<sup>3</sup>/4 instead of D<sup>3</sup>/6 for the second term and I just can't work it out, if anyone can help I'd appreciate it.</p>
| |mechanical-engineering|fluid-mechanics|chemical-engineering|process-engineering| | <p>For your curved surfaces, the lever arm for the calculation of torque is a constant $D/2$. That is not true in the case of the cavern's faces. In that case the lever arm is a variable ($r$). To calculate the torque you would need to integrate the stress over the area times the lever arm. It should look something like this:</p>
<p>$$
\int\int\tau_y\cdot r\, dA
$$</p>
<p>where the double integral is over the area where the force acts. Be careful to pick the appropriate differential area for a constant z surface in cylindrical coordinates. I did the calculation and the answer matches the one you state.</p>
| 8267 | Working Out Shaft Torque in a Stirred Tank with a Cylindrical Cavern Formation (using a non Newtonian fluid) |
2016-04-03T17:43:38.397 | <p>How does union joint with two male threaded ends work? Does one male end rotate 360 degrees around it axis? How is the inside of the union constructed? Can I connect two pipes with female threads using a union joint and assemble/disassemble it without too much pain? Specifically, I am looking at a stainless steel union joint.</p>
| |piping|pipelines|joining| | <p>Typically, most threads on pipes (in the USA at least) are male <a href="https://en.wikipedia.org/wiki/National_pipe_thread" rel="nofollow noreferrer">NPT</a> and all the couplings or fittings are female NPT.</p>
<p><a href="https://i.stack.imgur.com/hZoVe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hZoVe.jpg" alt="enter image description here"></a>
<br><br><br>
You can connect two pipes with a coupling as long as the last pipe added is free to rotate about its axis.</p>
<p><a href="https://i.stack.imgur.com/M6Axg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M6Axg.jpg" alt="enter image description here"></a>
<br><br><br>
A union is used for easy removal of pumps and other serviceable equipment that would otherwise not be easily unthreaded. It is also used to attach pipes in which neither pipe can not be rotated about its axis (perhaps because it is assembled to other plumbing, etc). The following image is the best section view of a union I could find. Most of the ones I have used have a flat steel face that mates to a flat steel face with an oring (with no 45 degree bevel like shown in the image). The flat faces allow the equipment to easily be removed sideways out of the plumbing when the union is detached.</p>
<p><a href="https://i.stack.imgur.com/JMM3K.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JMM3K.jpg" alt="enter image description here"></a></p>
<p><a href="http://www.smithcooper.com/catalogue/group/2781/nameblock/Union" rel="nofollow noreferrer">Stainless Steel Unions with datasheet links</a></p>
<p>Most hardware/plumbing stores sell pipe unions and will let you take them apart to see if they fit your needs. Seeing them first hand is the best way to get a feel for how they work.</p>
| 8272 | How does a union joint with two male threads work? |
2016-04-04T22:05:14.957 | <p>I'm trying to understand the derivation of the turbulent kinetic energy equation, as described in this link: <a href="http://www.diva-portal.org/smash/get/diva2:379743/FULLTEXT01.pdf" rel="nofollow">Evaluation of RANS turbulence models for flow problems with significant impact of boundary layers</a>.</p>
<p>I'm able to follow up to equation 2.26 on slide 11 (i.e. page 9), which states</p>
<p>$$\rho \overline{\frac{\partial u_i'}{\partial t}u_i'} + \rho\left( \overline{u_j\frac{\partial u_i}{\partial x_j}u_i} -\bar{u}_j\frac{\partial u_i'}{\partial x_j}\bar{u}_i\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i,$$</p>
<p>where the overbars denote the ensemble averages of the quantities under the bars. Note that here, the average of a product is not necessarily equal to the product of the average of each term. </p>
<p>According to the link, using only averaging rules, the term</p>
<p>$$\left( \overline{u_j\frac{\partial u_i}{\partial x_j}u_i} -\bar{u}_j\frac{\partial u_i'}{\partial x_j}\bar{u}_i\right)$$ </p>
<p>can be simplified to become</p>
<p>$$\overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \overline{\frac{\partial u_i'}{\partial x_j}u_j'}\bar{u}_i + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}.$$</p>
<p>If I substitute this directly into the previous equation, I get </p>
<p>$$\rho \left( \overline{\frac{\partial u_i'}{\partial t}u_i'} + \overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \underbrace{\overline{\frac{\partial u_i'}{\partial x_j}u_j'}\bar{u}_i}_{\text{my 4th term}} + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i.$$</p>
<p>However, their derivation yields</p>
<p>$$\rho \left( \overline{\frac{\partial u_i'}{\partial t}u_i'} + \overline{u_i'\frac{u_i'}{\partial x_j}}\bar{u}_j + \overline{\frac{\partial u_i'}{\partial x_j}u_i'u_j'} + \underbrace{\overline{\frac{\partial u_i'u_j'}{\partial x_j}}\bar{u}_i}_{\text{their 4th term}} + \overline{u_j'u_i'}\frac{\partial u_i}{\partial x_j}\right) = -\overline{\frac{\partial p'}{\partial x_j}u_i'} + \nu\overline{u_i'\nabla^2u_i'} + \rho \frac{\partial (\overline{u_i'u_j'})}{\partial x_j}\bar{u}_i.$$</p>
<p>The only difference between my derivation and theirs is the 4th term on the left hand side of the equation. I'm sure that their 4th term is correct, since it is supposed to cancel out with a term on the right hand side of the equation. However, I can't seem to figure out how they obtained their 4th term on the LHS of the equation. The link suggests that the chain rule and the incompressiblity assumption are involved, but I'm not sure how.</p>
<p>In particular, the overbar term $\overline{u_i'u_j'}$ is treated as a single term. Thus, how can I apply the chain rule to $\frac{\partial (\overline{u_i'u_j'})}{\partial x_j}$? </p>
<p>Any hints to complete the missing steps would be greatly appreciated.</p>
| |fluid-mechanics|turbulence| | <p>The two terms are equal due to the product rule and continuity equation. The use of the continuity equation might not be obvious, however.</p>
<p>In their notation, the Reynolds decomposition is $u_i = \overline{u}_i + u^\prime_i$. Taking the continuity equation and averaging leads to:</p>
<p>$$\frac{\partial \overline{u}_j}{\partial x_j} = 0$$</p>
<p>Subtract the equation above from the non-averaged continuity equation to see that the fluctuations also are divergence free:</p>
<p>$$\frac{\partial u^\prime_j}{\partial x_j} = 0$$</p>
<p>Now, apply the product rule to the term of interest (before averaging):</p>
<p>$$\frac{\partial u^\prime_i u^\prime_j}{\partial x_j} = u^\prime_j \frac{\partial u^\prime_i}{\partial x_j} + u^\prime_i \frac{\partial u^\prime_j}{\partial x_j}$$</p>
<p>The second term is zero by continuity as seen above, and this completes the derivation. The averages commute, so the average of a derivative is the same as the derivative of the average.</p>
| 8285 | Deriving the turbulent kinetic energy equation |
2016-04-05T08:42:50.100 | <p>In Autocad 2015 for Mac, I have created a cylinder with a torus wrapped around it. The torus is 2 mm thick, with 1 mm embedded inside the cylinder and 1 mm outside. I want to subtract the torus from the cylinder, so I will be left with a semicircular groove around the cylinder. I have already done this once, but I had to remake the object so I deleted the original and started again - and now when I try to subtract the torus, it simply deletes the torus and leaves the cylinder untouched. What is going wrong, and how can I fix it?</p>
| |design|autocad| | <p>To be honest this came entirely from chance and I still have NO idea what caused the problem. The answer by @ChrisJohns will probably be a more likely cause of the issue, as for me I'm simply going to put it down to a bug, because this is how I fixed it:</p>
<p>Turn it off and on again. Make your torus/etc and put it in place, but if subtract doesn't work, undo the subtraction, save, quit, reopen and try it again, one subtraction at a time. Like I said, no idea what the cause is, so if this still doesn't work take a look at @ChrisJohns answer</p>
| 8292 | How to subtract a region where two solids intersect? |
2016-04-05T12:12:48.140 | <p>Can anyone tell me how to go about this problem? </p>
<p>There is a square shaped laminate (polyethylene fiber and epoxy resin composite) that is constrained on one side. A gun is fired and the bullet hits the laminate. What kind of stresses result in the laminate? How should I go about their calculation? When I say stresses, I mean the stress tensor components $\sigma_{i,j}$. Also, can the strain tensor components $\epsilon_{i,j}$ be calculated?</p>
<p>Textbooks and other sources don't show or tell how the stress components are measured. Also, in the above problem, the mass and the velocity of the bullet are known.</p>
| |composite| | <p>An additional complication is the behaviour of the bullet itself on impact, which is not only difficult to model but can vary a lot according to the construction of the bullet, it's geometry and mass, and the angle and energy of impact. </p>
<p>For example a low-velocity, unjacketed lead bullet might squash against the impact surface with little or no penetration, which is a reasonable approximation of an inelastic collision and so the best modelling approach might be to consider its momentum and here it is certainly possible, although not trivial, to construct fairly credible finite element model based on dynamic stress and strain. Indeed you may even be able to get meaningful order of magnitude numbers with manual calculations. </p>
<p>On the other hand a high-velocity, armour-piercing bullet might well pass through cleanly and transfer relatively little energy to the target. </p>
<p>A third case is that fragmentation or tumbling of the bullet or target. In high energy impacts you may even observe fluid-type behaviour. </p>
<p>In addition to the actual contact surface between the bullet and the target there will also be shock waves that propagate through the target, which can cause high stresses and failures (eg. spalling) and can eject material from the back side of the target even if the bullet itself does not penetrate it. </p>
<p>In general the big difficulty with modeling this sort of situation is that you are dealing with strain rates far outside the assumptions underlying the material properties and equations used for static and most dynamic modelling. </p>
<p>To put it another way firing a bullet at a target is a fundamentally different condition from pushing it through a target slowly regardless of how much force is applied. Indeed in practical impact testing it is more usual to think in terms of energy than stress and strain. </p>
<p>In fact when it gets right down to real world applications standards for body armour will usually specify resistance to specific rounds (in terms of caliber, propellant and bullet type) at a specific range. Simply because there isn't a measurable physical property which is generally descriptive of this sort of performance. </p>
<p>You also need to think carefully about what you are trying to determine. In the context of armour something which generates a wide cone of debris and fragmentation on impact may actually be far worse than just allowing the bullet to pass through intact. </p>
| 8293 | Impact load of a bullet on a laminate |
2016-04-05T18:25:28.110 | <p>How do people design height gaining based on nested poles ?</p>
<p>Consider this picture (I made them) :
<a href="https://i.stack.imgur.com/JQUC2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JQUC2.png" alt="enter image description here"></a></p>
<p>If I want to achieve height based on these 3 poles, how they should behave?</p>
<p>Is this how the behave?</p>
<p><a href="https://i.stack.imgur.com/J9JdM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J9JdM.png" alt="enter image description here"></a></p>
<p>I had been searching over the internet I haven't found the related material, and it's kinda hard what to type. I'm doing some tasks on designing a simple machine (no foundation knowledge but with will) and my English is sort of limited, especially to technical stuff but I hope you understand it .</p>
<p>*Imagine like there's somehow an auto increasing portable stair to get over the tree, the poles(both sides) are consisted of poles fragment on each of them, how the fragments behave in this case. </p>
| |mechanical-engineering| | <p><em>Telescoping</em> is the English term for this and a search for <a href="https://www.google.com/search?q=telescoping%20mechanisms&source=lnms&tbm=isch&sa=X&gws_rd=cr#q=telescoping%20mechanisms&tbm=isch&pws=0&gws_rd=cr" rel="nofollow">telescoping mechanisms</a> will give lots of info on the various approaches.</p>
<p>A motor + transmission mechanism is called an actuator. So searching for <a href="https://www.google.com/search?q=telescoping%20actuator&biw=1280&bih=610&source=lnms&tbm=isch&sa=X" rel="nofollow">telescoping actuator</a> will show you devices that automatically extend with some sort of external or internal control.</p>
| 8302 | Principal on gaining more height based on nested poles? |
2016-04-06T05:07:05.793 | <p>Is there a big difference between salt water and normal tap water in terms of thermal properties (heat capacity, for example)? </p>
<p>If so, can anyone give me a table of thermal and thermodynamic properties of salt water? </p>
| |fluid-mechanics|thermodynamics|heat-transfer| | <p>MIT provide a table of water properties at a range of salinity and temperature values: <a href="http://web.mit.edu/seawater/" rel="nofollow">http://web.mit.edu/seawater/</a></p>
<p>There is also excel/matlab correlations as well.</p>
| 8311 | Thermal properties of salt water |
2016-04-06T10:41:47.753 | <p>Suppose there is a cubic material with an internal heat source ($\Delta q / \Delta t =$ Constant), and is immersed in a sufficiently large amount of water. Now I would like to use finite difference method to simulate the steady state temperature distribution.</p>
<p>I think a natural convection boundary condition within the range of laminar flow can be assumed, and the maximum temperature should not exceed 100 $^\circ$C. As the surface temperature distribution is not uniform, each surface node should have its unique convective heat transfer coefficient ($h$), as this parameter depends on the temperature difference between the solid wall and the fluid.</p>
<p>But by looking up some heat transfer textbooks, I have never seen people address to the problem in such a way. I only found some treatments and correlation equations on the scenario of uniform temperature distribution of different geometries, while in my simulation, the temperature of different surface nodes are different. Giving it a second thought, I realized that convection heat transfer coefficient is dependent upon fluid properties, which will make it extremely complicated.</p>
<p>So is it possible to perform such calculations with finite difference method? Is there a method to at least estimate the convective heat transfer coefficient?</p>
| |thermodynamics|heat-transfer|chemical-engineering|thermal-conduction|convection| | <p>Your approach is absolutely acceptable so long as you realize its limits. The reason you are not seeing a method like this in a text book is because it is a mix of numerical methods and conventional empirical methods that is not particularly common. </p>
<p>The reason it is not common to approach a problem this way is because the natural convection heat transfer coefficient (htc) is a bulk property. It is based on the overall geometry and is going to give you a prediction of the heat transfer over the entire surface of the object. Thus the htc is based on the overall system geometry and properties and not the local geometry and properties right around each node. My hope is this image will make that clear:</p>
<p><a href="https://i.stack.imgur.com/swula.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/swula.png" alt="enter image description here"></a></p>
<p>Here we have a hot plate in some cold fluid. We have broken the hot plate into 6 nodes to estimate the surface temperature, but we can't calculate the htc based on only each nodes temperature because that is not the way natural convection works. </p>
<p>The most robust way to solve this is to solve the natural convection numerically as well. This is what a commercial CFD package like comsol would do. However, this would require that you model the fluid numerically which as you say above would get very complicated. </p>
<p>The other approach you could take is to continue using the bulk htc but recalculate it each time you iterate through your internal numerics. Then you would come up with a bulk htc and you could calculate the heat loss from each node via <code>Q=dA*htc*(T_i-T_bulk)</code>. When you are running your solver you just need to recalculate the htc based on a new average surface temperature each iteration. </p>
<p>There are other compromises you could make here to increase the accuracy of your model, but the overarching theme is to know the limits of each compromise you make. Using a bulk htc is an ok move if your fluid has a high Cp and thus will not have a big dT as it flows over the cube. It is also an ok move if the geometry is a cube or a sphere where the water is not flowing over the part for a long time. This would be a bad approach if your fluid had a low Cp and thus had a big dT as it flowed over your part or if the part had a very long flow path as it does not capture the change in duty caused by the change in dT between the first and last node the water flows past. </p>
<p>All in all, I think the model is simple enough you should just write it and see what the results are. You can then start messing with the temperature you use for the bulk temperature to try to capture the water's temperature better. </p>
| 8315 | Convective heat transfer coefficient in finite difference method boundary conditions |
2016-04-06T13:07:24.357 | <p>I have about 12 bottles of water. If I want to minimize the energy consumption of my refrigerator, should I put these bottles in the fridge or leave them out?</p>
| |thermodynamics|energy-efficiency|refrigeration| | <p>Many factors play in this scenario:<br>
1- the shape of the bottles and their material thermal insulation index.<br>
2-how you place the bottles on the shelves: do they promote air current or impede it.<br>
3- the frequency of opening the fridge door.<br>
As many have mentioned there is an initial load when room temperature bottles cool of. but after that point everything else being equal it boils down two cases; penetration of heat into fridge due to design and manufacturing exigencies and material thermal properties of fridge without bottles versus contribution of bottles acting like the vanes of a cold radiator letting the stream of transient air wash around it. So if the shape and placement of bottles doesn't encourage entrant air circulate easily it would help saving energy a bit.<br>
ultimately if you fill up the entire available space with water past the initial pick of energy the chronic use of energy will be smaller.<br>
we could look at this fridge as revers of a cooling tower. if we don't want to interact thermally with ambient air we make it hard by placing the bottles of water in a way that shields the other parts of the fridge from convection or ventilation. this concept has been used for centuries in middle eastern architecture, eg, passive air-conditioning.</p>
| 8316 | Difference in energy consumption between empty fridge and fridge filled with bottles of water |
2016-04-06T23:33:31.797 | <p>I'm trying to calculate the clamping force resulting from torquing a nut and bolt to a particular level.</p>
<p>I have found this formula in various forms in a lot of places.</p>
<p>$$T = KDP$$</p>
<ul>
<li>$T$ = Torque (in-lb) </li>
<li>$K$ = Constant to account for friction (0.15 - 0.2 for these units)</li>
<li>$D$ = Bolt diameter (inches)</li>
<li>$P$ = Clamping Force (lb)</li>
</ul>
<p>I applied this to my problem</p>
<ul>
<li>$T = 0.6\text{ N-m} = 5.3\text{ in-lb}$</li>
<li>$D = 3\text{ mm} = 0.12\text{ in}$</li>
<li>$K = 0.2$</li>
</ul>
<p>This gives $P = \dfrac{T}{KD} = 220\text{ lb} = 100\text{ kg}$.</p>
<p>So, I have two questions.</p>
<ul>
<li>The result seems way too high. I'm using a tiny M3 bolt, and not much torque. I can't see how this would result in 100 kg force. Can anyone see the error?</li>
<li>The formula doesn't take account of thread pitch. I would expect a fine thread to give more clamping force for the same torque. Is there a formula that accounts for thread pitch?</li>
</ul>
| |mechanical-engineering|applied-mechanics|torque| | <p>The correct answer is 100 N-m (force) not 100 KG (mass).</p>
<p>The error was in failing to keep all the units consistent throughout.</p>
<p>Off the top of my head 1 N force is about the force exerted by gravity for one "standard apple." That equates to about 0.25 pounds.</p>
| 8324 | Calculation of Clamping force from bolt torque |
2016-04-07T12:20:47.343 | <p>I am measuring the rotational speed of a shaft using a disk with 60 cuts and an optical fork. The signal from the optical fork is sent to a frequency to voltage converter and after that to a data acquisition system.</p>
<p>What is the working principle of this frequency to voltage converter? Is it counting pulses?</p>
| |measurements| | <p>There are a number of ways.</p>
<p>As Russell described, one way is to make a fixed size pulse each period, then low pass filter the result.</p>
<p>Another is to make a frequency-dependent filter. Feed in a signal with fixed amplitude and varying frequency. The filter varies the amplitude as a function of frequency. You then make a AC to DC amplitude converter, often called a "detector". Such a detector is at the heart of a AM radio, for example.</p>
<p>You can have a bunch of narrowly tuned circuits in parallel, then see which one has the highest response to the input signal.</p>
<p>You can sample the input signal, convert it to digital values, and run a FFT (fast fourier transform) on them.</p>
<p>If you are expecting only a single input frequency, you can tweak a oscillator to attempt to mimic that frequency. If the oscillator is voltage-controlled and you do this in a feedback loop, then the control signal into the oscillator becomes the voltage signal indicating the input frequency. This may sound far fetched, but this method is common enough to have a name. It's called a <i>phase locked loop</i> (PLL), and you can get PLL chips that do most of this for you. Most newer (last 40 years or so) FM radios work on this principle.</p>
<p>Old FM radios used a different technique, called a <i>ratio detector</i>.</p>
| 8333 | How does a frequency to voltage converter work? |
2016-04-08T17:48:51.833 | <p>I'm finding myself using control charts and I'm unclear on a very basic point. I've read page after page of texts, but they all focus on how to make the initial chart rather than walking you through how to update it.</p>
<p>To make things easier, I'll set the situation:
I'm making an Individual Control Chart. Every thirty points, the chart is updated. In January, I put points 1-30 on the chart. February, I put points 31-45 on, and March points 45-60. So, the chart needs to be updated now. Where do I go from here?</p>
<p>Assuming none of the data points were out of control, here's what I think happens:
I think points 1-30 are removed from the chart. I think the chart has the average and control limits calculated using data points 31-60. I think that future points (61-90) are plotted on the newly updated chart until data point 90 is reached, at which point the chart is updated, points 31-60 dropped, and the average and limits recalculated using points 61-90.</p>
<p>I am aware that the limits for updated versions of the chart should be calculated using xbarr +/- 3s/c4. </p>
<p>Edit:</p>
<p>So, using the power of randbetween, I've made some sample charts and sample data. The first is for the month of January, where data was taken every day, giving 31 data points, and at the end of the month, the chart that follows.</p>
<p>Date Value Average SD UCL LCL
1/1/2016 9 6.161290323 2.887868582 14.82489607 -2.502315423</p>
<p>And a bunch of values for the dates stretching out over the next few months (I can post all 60 rows if really desired).</p>
<p><a href="https://i.stack.imgur.com/axyOd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/axyOd.jpg" alt="sample_chart_01"></a></p>
<p>Okay, so I've plotted the first 30 datapoints. So far so good. In February, only 15 samples are sampled for data. The datapoints are added to the chart for a total of 45 datapoints on the chart:
<a href="https://i.stack.imgur.com/jpN71.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jpN71.jpg" alt="sample_chart_02"></a></p>
<p>In March, another 15 samples are sampled and this brings the total number of datapoints up to 60:</p>
<p><a href="https://i.stack.imgur.com/YEZQb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YEZQb.jpg" alt="sample_chart_03"></a></p>
<p>Now at this point, I know that the chart must be updated. But there are a few ways I can do it. One of which is to remove the January data (the first 30 datapoints) both from the chart and from the calculations and only plot datapoints 31-60, with datapoints 61-90 being plotted on this as well (at which time, we would go through the process again):
<a href="https://i.stack.imgur.com/XaqvF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XaqvF.jpg" alt="sample_chart_04"></a></p>
<p>Or I could calculate the average, SD, and the upper and lower limits based on datapoints 31-60, but plot datapoints 1-60:
<a href="https://i.stack.imgur.com/Ev6s8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ev6s8.jpg" alt="sample_chart_05"></a></p>
<p>Or I could re-calculate the average, SD, and upper and lower limits on datapoints 1-60 AND plot datapoints 1-60:
<a href="https://i.stack.imgur.com/sIHcM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sIHcM.jpg" alt="enter image description here"></a></p>
<p>All of these give different limits, different charts, and all of them seem equally valid because nowhere does it seem to discuss what you do to actually <em>update</em> the chart once it has been made.</p>
| |statistics|quality-engineering| | <p>Out of desperation, I got in contact with some very nice statisticians at NIST. They walked me through the answer to my question, which boils down to: control charts are an empirically derived chart with strong statistical backing. However, because of this by its very nature, there is no one right answer. It depends on how historically stable your process is, whether or not new changes need to be accounted for, etc. You have to have an understanding of your data, an understanding of what is relevant and how that will affect your data in order to correctly decide. It's a judgement call based on someone who understands the process enough to be able to call when the limits and such are clearly incorrect and must be done a different way. Hope this helps anybody in the future needed help on control charts.</p>
| 8349 | How do I update an individual control chart? |
2016-04-08T21:15:48.993 | <p>Does anyone know how to determine the moment of inertia of the filled circular sector shown below?</p>
<p><a href="https://i.stack.imgur.com/sqZ9I.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sqZ9I.jpg" alt="enter image description here"></a></p>
<p>Apparently I have to use $dA=r\ d\theta\ dr$, $y= r\sin(\theta)$ and end up getting $I_x=\frac{r^4}{8}(\alpha-\sin(\alpha))$</p>
<p>But I have no idea how they got that, I have never had this.
Could someone step by step explain how to get to $I_x$ with this question and how to integrate something that has two different differentials.</p>
| |statics|mathematics|moments| | <p>Consider an infinitesimal element of area $r d\theta dr$ which is at a distance $r \sin (\theta)$ from the $x$ axis. </p>
<p><a href="https://i.stack.imgur.com/d9NGX.jpg"><img src="https://i.stack.imgur.com/d9NGX.jpg" alt="enter image description here"></a></p>
<p>Its moment of inertia is $r d\theta dr (r \sin (\theta ))^2$.</p>
<p>The moment of inertia about the $x$ axis of the complete sector:</p>
<p>$$ I_x = \int _0^{r_0}\int _{-\frac{\alpha }{2}}^{\frac{\alpha
}{2}}r^3\sin ^2(\theta) d \theta dr = \frac{1}{8} r_0^4 (\alpha -\sin (\alpha ))$$</p>
| 8355 | Determine the moment of inertia of a filled circular sector |
2016-04-09T18:30:49.700 | <p>What is the physical meaning of using negative frequencies while making Nyquist plots?</p>
<p>I know that we do a mapping from $s-$ plane to $f(s)$ plane, and we move along a contour. What I don't get is that Is the Nyquist plot purely a mathematical construct? Is there no relation between the frequency on plot and actual frequency? If there is, then what is the point of plotting negative frequencies,Is it just so that we could count the number of encirclements?</p>
<p>I am not able to grasp the idea behind it.</p>
| |control-theory| | <p>The negative frequencies are not physical, neither do they convey any new information. The value at the negative frequency is the complex conjugate, and hence the plot is symmetric about the real axis. If you know one you can get the other as a conjugate.</p>
<p>The value in plotting the negative values (or going around the entire contour) is in doing the stability analysis. </p>
| 8363 | Why are negative frequencies included in the nyquist plot? |
2016-04-10T07:54:08.047 | <p>I'm an engineering enthusiast but I don't yet have a formal education in engineering and I'm hoping you guys can tell me about the implications of using a metal road design in terms of durability and cost. </p>
<p>My (vague) idea is (for 2 lanes of road) a set of 3 rails laid down as a vertical layout for a length of road, similar to railroad tracks, which supply power for a self-deploying set if metal road panels. The horizontal panels are designed to drain water and grooved for tire traction. The way the panels self deploy isn't really important, it's just a mechanism that rides on the three rails and lays down a load of the metal panels.</p>
<p>The self deployment isn't really important it's just to lower the construction cost and explain the 3 rail design I had in mind. </p>
<p>I'm purposely being vague when I say metal, expecting answers to provide the kind of metal that would be optimal because I don't know. </p>
<p>It's just that we spend billions across the globe replacing roads, and while metal would obviously cost a lot more than cement, I'm really curious about whether the world would save money by building very durable roads which need to be replaced less often and which for the most part might be replaced systematically the same way they're deployed, requiring only a small fraction of the maintenance of a traditional road replacement. </p>
<p>Obviously the real maintenance cost of this imaginary design is unknown, my question is purely about the difference in cost for materials and the durability so I can have some basis for speculation about the viability of said design. </p>
| |civil-engineering|highway-engineering|roadway| | <p>Asphalt, which is essentially crushed aggregate bound with bitumen has a lot of advantages for road building. </p>
<ul>
<li>It is porous and so can drain by allowing water to pass through it as well as running off the surface. </li>
<li>Because it is made up of particles of crushed masonry it has good friction characteristics in all directions (ie you get the same grip
turning as you do braking) which isn't greatly affected by wear. ie the 'grip' is largely inherent in the material rather than depending on a particular surface coating or treatment. </li>
<li>Asphalt is pretty easy to lay as it can be continuously poured onto a prepared subsurface either by specialist plant or just by shoveling out out of the back of a flatbed truck. </li>
<li>In road building it is the foundation layers of the road which are the expensive bit and even then the cost is largely associated with the planning, design, surveying and infrastructure (utilities, signage, bridges, culverts, drainage etc) rather than the materials. </li>
<li>It can be repaired reasonably easily by planing off the top layer and relaying or even adding a dressing layer over the old surface.</li>
<li>properly made repairs should be more or less seamless and permanent.</li>
<li>it is possible to lay lower grade temporary surfaces during major works quickly and cheaply. </li>
<li>serviceable temporary repairs for potholes etc can be made very cheaply</li>
<li>Trenches for utilities etc are easy to dig and fill in. </li>
<li>Because the binder is relatively soft small defects can self-repair, especially in hot weather. </li>
<li>Expansion joints are not required. </li>
<li>Asphalt can be laid over an uneven surface and doesn't require precise leveling as it doesn't have discrete joints. </li>
<li>Deterioration of asphalt roads over time is inconvenient but it is unlikely to lead to catastrophic failures in contrast to eg steel
railway lines. </li>
<li>Asphalt isn't subject to much in the way of corrosion or environmental degradation.</li>
</ul>
<p>There are some very obvious potential problems with a road with a steel deck</p>
<ul>
<li>The surface would need to be textured to provide grip and water drainage. This would be ground away fairly rapidly by dirt and grit carried onto the road and the action of tyres and so would need renewal. </li>
<li>Joints between sections would need to be well aligned to provide an adequate surface. </li>
<li>How would they be joined ? Steel structures are generally either bolted or welded but both are expensive on this sort of scale compared to just raking out hot asphalt. </li>
<li>You would still need to prepare a stable substructure. </li>
<li>The construction would need to accommodate thermal expansion and expansion joints in bridges etc add a lot of noise and vibration when driven over. </li>
<li>even mild steel is moderately expensive and requires a lot of energy to produce. </li>
<li>Most structural steels need protection from corrosion by painting, coatings etc which require regular maintenance when exposed to the elements. This will be seriously exacerbated in regions where roads need to be salted in the winter. </li>
</ul>
<p>For a real world comparison we can look at the <a href="http://www.transport-watch.co.uk/facts-sheet-7-rail-versus-road-capital-costs-track" rel="nofollow">relative cost per km</a> of railway track vs multi-lane roads. This puts rail at around 6 times more expensive per km than roads. Note also that a large proportion of the cost of these projects is associated with structures such as bridges and signs/signalling as well as the cost of the land itself, rather than the actual road surface. </p>
<p>See also <a href="http://www.dot.ca.gov/hq/esc/oe/cost_index/historical_reports/PI_2012_qtr4.pdf" rel="nofollow">this document</a> for estimates of civil engineering materials by weight. Note that it puts <strong>asphalt as 94 dollars per ton</strong> and <strong>structural steel at 2586 dollars per ton.</strong> This doesn't necessarily tell you a lot about the total costs of what would necessarily by very different approaches but illustrates that asphalt is a very inexpensive bulk material. </p>
<p>It is also worth considering that even in the case of steel bridges the road surface is almost invariably asphalt or concrete except where the structure is temporary or take very low traffic. </p>
| 8367 | What would be the implications of a metal highway design? |
2016-04-12T01:49:34.647 | <p>Why does the AISC handbook only list one figure for Section Modulus for shapes such as Angle(L) steel? Shouldn't there be two? The steel is asymmetric. Is the figure given just the Delta to the bottom or something; figuring a design only needs to be concerned with Tension?</p>
| |steel|geometry| | <p>To quote <a href="https://en.wikipedia.org/wiki/Section_modulus" rel="nofollow">Wikipedia</a>:</p>
<blockquote>
<p>The elastic section modulus is defined as $S = \dfrac{I}{y}$, where $I$ is the second moment of area (or moment of inertia) and $y$ is the distance from the neutral axis to any given fibre. <strong>It is often reported using $y = c$, where $c$ is the distance from the neutral axis to the most extreme fibre</strong></p>
</blockquote>
<p>The tables only show the section modulus for the fiber farthest from the centroid. <a href="http://www.engineersedge.com/materials/aisc_structural_shapes/aisc_structural_shapes_viewer.htm" rel="nofollow">This properties viewer</a> shows L8x8x1-1/8 with $d = b = 8 \text{ in}$, $\overline y = 2.4 \text{ in}$ (from the horizontal leg), $I = 98.1 \text{ in}^4$, and $S = 17.50 \text{ in}^3$.</p>
<p>$$\begin{align}
S_{top} &= \dfrac{I}{d-\overline y} \approx 17.50 \\
S_{bot} &= \dfrac{I}{\overline y} = 40.875 \\
\end{align}$$</p>
<p>One can easily get from one section modulus to another, though (but it might just be easier to calculate it with the traditional equations):
$$S_{bot} = S_{top}\dfrac{d-\overline y}{\overline y}$$</p>
<p>It is worth stating however that the critical fiber may not be the one farthest from the centroid. While the farthest face will always have the largest (in magnitude) stress, the near face may be the controlling factor in design. After all, it may be that the lower compressive stress on the near face may lead to buckling before the higher tensile stress on the far face reaches the material's ultimate strength.</p>
| 8391 | AISC Section Modulus numbers |
2016-04-12T15:34:23.993 | <p>The two terms seem interchangeable and I can't find a strict definition for both, so what's the difference between them if there is any?</p>
| |mechanical-engineering|pumps|turbomachinery| | <p>A centrifugal pump consists of two main components</p>
<ol>
<li><p><strong>Electric motor</strong><br />
It is usually an induction motor. All electrical engineering courses would cover this in the lower year.<br />
An electric motor has two components</p>
<ul>
<li><strong>Stator</strong> - It is the stationary part of the motor. Electricity is supplied to terminals on the stator.</li>
<li><strong>Rotor</strong> - It is the rotational part of the motor. It is supported on bearings and mechanically coupled to external device, such as impeller in this case.</li>
</ul>
</li>
<li><p><strong>Impeller</strong><br />
This is the rotational part of a centrifugal pump or a compressor. It is used to increase the pressure and flow of a fluid. As it is coupled to the rotor of a motor and rotational, it is also called <em>rotor</em>.</p>
</li>
</ol>
<p>Induction motor has very limited speed control, hence <strong>variable speed drive (VSD)</strong> is used to supply variable frequency and voltage to the motor, enabling the later to run at different speed, and hence driving the impeller to operate at different pressure and flow rate.</p>
| 8398 | What's the difference between impeller and rotor in centrifugal pump? |
2016-04-13T11:58:42.207 | <p>I am trying to solve the problem:</p>
<blockquote>
<p>The critical resolved shear stress for iron is 27 MPa. Determine the
maximum possible yield strength for a single crystal of Fe pulled in
tension.</p>
</blockquote>
<p>I think that this is related to the equation $\sigma_{yield} = \dfrac{\tau_{crit}}{\cos{\Phi}\cos{\gamma}}$.</p>
<p>The book says:</p>
<blockquote>
<p>The minimum stress necessary to introduce yielding occurs when a
single crystal is oriented such that $\Phi=\gamma=45°$; under these
conditions, $\sigma_{yield}= 2\tau_{crit}$.</p>
</blockquote>
<p>When I tried check my answer, the solution manual said I should use $\sigma_{yield}= 2\tau_{crit}$:</p>
<blockquote>
<p>In order to determine the maximum possible yield strength for a single
crystal of Fe pulled in tension, we simply employ Equation 7.5 as
$\sigma_y=2\tau_{crss}=(2)(27\ \mathrm{MPa})=54\ \mathrm{MPa}$.</p>
</blockquote>
<p>I don't get this. Doesn't this calculate the minimum yield strength, not the maximum? The solution manual is considering the possibility of the iron crystal being pulled in tension in a direction that is most favourable and makes one of the slip systems reach the critical value the quickest.</p>
<p>But to know the maximum possible yield strength shouldn't you calculate the situation in which tension is applied in a direction that makes $\cos\Phi\cos\gamma$ as small as possible? So, basically, in a direction in which the resolved shear stress is just a fraction of the applied tensile stress?</p>
<p>The solution manual gets the yield strength when you pull the single iron crystal in such a direction that the resolved shear stress is the largest fraction possible of the applied stress. But if you reorient the single crystal, couldn't it withstand a way larger applied tensile stress and thus have a higher yield strength?</p>
| |mechanical-engineering|materials| | <p><strong>The book should probably have said "minimum possible yield strength" or "maximum possible resolved shear stress." Note that the maximum yield strength of any real single crystal does not depend on one slip system and thus can not be infinite despite what <a href="https://en.wikipedia.org/wiki/Schmid's_Law" rel="nofollow">Schmid's Law</a> might imply. The maximum yield strength occurs at a set of angles "furthest" from all of the slip systems and is more difficult to calculate.</strong></p>
<p>There appears to be confusion about what the Schmid's Law actually means. Schmid's Law relates applied tensile stress to resolved shear stress. In other words, a way of transforming an applied tensile stress to a shear stress in a given plane and along a given direction within that plane, based on the angles of the plane and direction with tension.</p>
<p>The yield strength of a single crystal is related, by Schmid's Law, to the critical resolved shear stress necessary to cause slip. The shear stress necessary to cause slip for a given slip system is a material property. The applied tensile stress on a single crystal is related, also by Schmid's Law, to the resolved shear stress on a given slip system, via the angles between slip plane and tension direction, and slip direction and tension direction. Slip plane and slip direction together make a slip system. Note that yield strength (a material property) is related to critical stress (a material property), and applied stress is related to resolved stress. When the applied stress is such that the resolved shear stress equals the critical resolved shear stress, then the applied stress is equal to the yield strength, and the crystal slips.</p>
<p>To determine the yield strength of a single crystal in a given direction, first pick a slip system (plane and direction), then pick the direction of tension. Finally determine what tensile stress is required to cause the actual resolved shear stress to equal the critical resolved shear stress. Schmid's Law allows you to convert tensile stress to actual resolved shear stress. Then replace actual resolved shear stress with critical resolved shear stress, and work backward to determine the yield strength.</p>
<p>To answer your other question, yes it is possible for the yield strength to vary depending on the choice of slip system and tension direction. Single crystals have anisotropic mechanical properties. Consider a (not-real) system with only one slip plane. If that one slip plane is oriented perpendicular or parallel to the direction of tension, slip is impossible because the actual resolved shear stress is 0 in both cases no matter what amount of tension is applied. In a sense, the theoretical tensile yield strength <em>by slip</em> of such a non-physical system is infinite in those directions.</p>
<p>All real crystal systems have slip systems such that slip can occur in every tensile direction, though the tensile strength may vary based on orientation. Which slip system becomes active depends on which system's resolved shear stress exceeds its critical resolved shear stress first, which in turn depends on the relative angles the systems make with the choice of tension direction. It is also important to note many crystalline materials behave isotropically in bulk due to individual grains being oriented randomly, unlike in a single crystal which has a dominant orientation.</p>
| 8407 | Determine the maximum possible yield strength for a single crystal of Fe pulled in tension |
2016-04-13T12:32:50.223 | <p>I know that an hydraulic jump can occur only when the Froude number is higher than 1 (supercritical flow), and I am familiar with the Bélanger equation. However, I feel like I have a fundamental misunderstanding about hydraulic jumps: if a jump can occur when Froude number is larger then 1, what prevents the flow in rivers from "jumping" immediately whenever its Froude number becomes slightly more than 1? </p>
<p>I mean how can a fluid develop a very high Froude number without making a hydraulic jump? According to what I read, in industrial applications such jumps are specially induced on the fluid to dissipate its energy. So I want to understand how it's done. </p>
| |fluid-mechanics|hydraulics|open-channel-flow| | <p>First a hydraulic jump is where fluid will transition from supercritical (having a Froude Number greater than one, to subcritical (having a Froude Number less than one). When the Froude number is near one, the jump is very weak and somewhat gradual such that you might not even be able to tell it is "jumping".</p>
<p>To figure out where the jump occurs you need to know when the Froude Number will transition from less than one to more than one. To help with this is an extremely useful equation for modeling open channel flow:</p>
<p><span class="math-container">$$\frac{dy}{dx}=S\frac{1-\left(\frac{y_n}y\right)^3}{1-\left(\frac{y_c}y\right)^3}$$</span></p>
<p>Where <span class="math-container">$y$</span> represents the depth of flow in a riverbed, x is the distance along the flow, <span class="math-container">$S$</span> is the slope of the riverbed (positive being flowing downhill, negative, flowing uphill), <span class="math-container">$y_c$</span> is the depth of flow corresponding to <span class="math-container">$Fr=1$</span> and <span class="math-container">$y_n$</span> is the <em>normal depth</em>, the depth of flow corresponding to flow that is balancing the frictional losses with gravitation gains.
<span class="math-container">$$y_c=\sqrt[3]{\frac{q^2}g}$$</span>
<span class="math-container">$$y_n=\sqrt[3]{\frac{q^2}{C^2S}}$$</span></p>
<p>Where <span class="math-container">$q$</span> is the flow rate per width of river, <span class="math-container">$C$</span> is Chezy's coefficient (approximated as constant), and <span class="math-container">$g$</span> is acceleration due to gravity.</p>
<p>A depth greater than <span class="math-container">$y_c$</span> corresponds to a slower velocity, and thus subcritical flow (having a Froude number less than 1). When this is the case, the bottom of the fraction is positive between 0 and 1.</p>
<p>Similarly, when the depth is less than <span class="math-container">$y_c$</span> the flow is supercritical and the bottom of the fraction is negative.</p>
<p>In this range therefore, if the top of the fraction is also negative, then the flow will be getting deeper, but if it is positive, the flow will be getting shallower. This corresponds to a depth of flow less than and greater than <span class="math-container">$y_n$</span> respectively. This means that while the flow is supercritical, the depth of flow will always be moving towards <span class="math-container">$y_n$</span></p>
<p>This means that if you have a steep slope such that <span class="math-container">$y_n$</span> is less than <span class="math-container">$y_c$</span> the depth of flow will move to and then stay at <span class="math-container">$y_n$</span>. If you would like the flow to have a hydraulic jump simply decrease the slope till <span class="math-container">$y_n$</span> is greater than <span class="math-container">$y_c$</span> then the flow will deepen at progressively faster rates until you reach the hydraulic jump.</p>
<p>Note since <span class="math-container">$q$</span> is based on your width it's also possible to modify the width rather than slope to cause <span class="math-container">$y_n$</span> to be greater than <span class="math-container">$y_c$</span>.</p>
<p>But wait! After the hydraulic jump the flow will be subcritical, which means the flow downstream can now influence the water level and push the hydraulic jump upstream. How far upstream? Depends on the flow downstream.</p>
<p>In subcritical flows it is easiest to work upstream. If you try to work downstream from an initial guess, you'll end up with non-physical flows unless you happen to guess perfectly. This is because in subcritical flows, as you move downstream, the depth moves <em>away</em> from the normal depth. If you're less than the normal depth, you can quickly end up with a negative depth, and if you're more than the normal depth you can end up with a depth the would overflow your walls quick quickly. However, if you work backwards, going upstream then the depth always moves towards the normal depth, which should always give a valid depth.</p>
<p>So then the question is where to get your initial condition? Downstream of subcritical flow, and upstream on subscrital flow. The opposite transition of hydraulic jump. This transition is always smooth and thus according to the equation always happens when <span class="math-container">$y=y_n=y_c$</span> with <span class="math-container">$y_n$</span> decreasing downstream. An extreme example would be the lip of a waterfall or edge of an overflowing dam. For given flow rate, the depth will always be easy to predict at these choke points. As an added bonus the slope at these choke points is largely flow rate independent:</p>
<p><span class="math-container">$$y_c=y_n$$</span></p>
<p><span class="math-container">$$\sqrt[3]{\frac{q^2}g}=\sqrt[3]{\frac{q^2}{C^2S}}$$</span>
<span class="math-container">$$\frac1g=\frac1{C^2S}$$</span>
<span class="math-container">$$S=\frac{g}{C^2}$$</span></p>
<p>So anytime the slope goes from shallower than <span class="math-container">$\frac{g}{C^2}$</span> to steeper there might be a transition to supercritical flow right there, and anytime the slope goes from steeper than <span class="math-container">$\frac{g}{C^2}$</span> to shallower there might be a hydraulic jump a bit upstream.</p>
<p>I say might because it is possible that a choke point further downstream is causing the depth to be too high to cause any transitions (you could be at the bottom of a lake for example.)</p>
<p>However, if the slope is constant and shallower than <span class="math-container">$\frac{g}{C^2}$</span> for a long time (many miles), then odds are, near the top of this section the depth will have converged to <span class="math-container">$y_n$</span>. So by this shortcut you don't have to start all the way back from a choke point.</p>
<p>In either case, you can model the flow going upstream from the subcritical section and downstream from the supercritical section. Then having modeled the supercritical section you can plot vs position the depth of flow that would result from a hydraulic jump at that location. Any time that plot intersects the subcritical plot is a location where a hydraulic jump could be stable.</p>
<p><sup>Equations are derived in full in <a href="https://ocw.mit.edu/courses/earth-atmospheric-and-planetary-sciences/12-090-introduction-to-fluid-motions-sediment-transport-and-current-generated-sedimentary-structures-fall-2006/course-textbook/ch5.pdf" rel="nofollow noreferrer">MIT's open courseware chapter on open channels</a>. My analysis of those equations was also guided by the analysis presented there.</sup></p>
| 8408 | Where exactly will a hydraulic jump occur? |
2016-04-13T12:34:26.537 | <p>I have trasfer functions of a plant and a controller in laplace domain. I checked for the closed loop response by applying a step response. The system is found to be stable. </p>
<p>I checked the response of the system for the same step reference in discrete domain. I can see that the closed loop response of the discrete system is unstable. The transfer function is converted to discrete form by c2d option in matlab with a sampling time of 1ms.</p>
<p>Shouldnt the response of the system be same in both continuous and discrete domain (atleast for high sampling frequency)?</p>
| |control-engineering|transfer-function|matlab|feedback-loop| | <p>The plant and controller:
$$\text{sys}=\frac{4700 s^2+4393 s+3.245\times 10^8}{s^4+7.574 s^3+120200. s^2}$$</p>
<p>$$pid=0.287\, +0.008 s+\frac{0.5}{s}$$</p>
<p>The closed-loop system obtained as $\frac{pid*sys}{1+pid*sys}$:</p>
<p>$$csys=\frac{37.6 s^4+1384.04 s^3+2.59961\times 10^6 s^2+9.31337\times 10^7 s+1.6225\times
10^8}{1. s^5+45.174 s^4+121584. s^3+2.59961\times 10^6 s^2+9.31337\times 10^7
s+1.6225\times 10^8}$$</p>
<p>The poles are all in the left-hand plane:</p>
<p>$$ \{-11.8643\pm \, 346.642 i,-9.80897\pm \, 25.3345 i,-1.82737\}$$</p>
<p>Thus as expected, the response to a unit-step is stable:</p>
<p><a href="https://i.stack.imgur.com/oB32V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oB32V.png" alt="enter image description here"></a></p>
<p>The zero-order hold approximation for a sampling period of 1 ms:
$$ \frac{-0.0371234 z^4+0.144584 z^3-0.213662 z^2+0.141976 z-0.0357747}{-1. z^5+4.83687
z^4-9.46905 z^3+9.38322 z^2-4.70688 z+0.955831}$$</p>
<p>The poles are all within the unit-circle:</p>
<p>$$ \{0.929426\pm \, 0.335734 i,0.989921\pm \, 0.0250846 i,0.998174\}$$</p>
<p>Again, as expected, the response to a unit-step is stable (and the plot is essentially the same):</p>
<p><a href="https://i.stack.imgur.com/L8bev.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L8bev.png" alt="enter image description here"></a></p>
| 8409 | Closed loop response of a discrete system |
2016-04-14T15:17:09.750 | <p>Given a nonlinear system, such as:</p>
<p>$$\begin{align}
x_1' &= x_2 \\
x_2' &= −x_1^3 + u \\
y &= x_2
\end{align}$$</p>
<p>How can I check the <strong>zero-state observability</strong> of the system?</p>
| |control-engineering|control-theory| | <p>I've found the answer.</p>
<p>To check if a system is <strong>zero state observable</strong>, put $u=0$ and check whether $x=0$ when $y=0$. If yes, it is <strong>zero-state observable</strong>. Otherwise not!</p>
<p>For the given system, by putting $u=0$ and $y=0$, we see that $x_2=0$, therefore $x'_2=0$ and thus $-x_1^3=0$ or $x_1=0 \implies x=0$. Thus it is <strong>zero-state observable</strong>.</p>
| 8423 | How can I check whether a nonlinear system is zero-state observable? |
2016-04-15T06:20:07.400 | <p>I am trying to perform simple 2D CFD analysis on Fluent having 3 adjacent zones(inlet, porous and outlet).
<a href="https://i.stack.imgur.com/YawGb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YawGb.png" alt="Zones figure" /></a></p>
<p>The problem I have is whenever I select non-equilibrium thermal model in porous zone in <em>Cell Zone Conditions</em> (as follows)
<a href="https://i.stack.imgur.com/dBto0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dBto0.png" alt="enter image description here" /></a></p>
<p>I get the following error:</p>
<blockquote>
<p>A duplicate of this zone cannot be generated automatically, as it
contains a non-conformal mesh interface.</p>
<p>You must manually make a copy
the porous fluid zone and define it as a solid zone before you enable
the non-equilibrium thermal model.</p>
</blockquote>
<p>AFAIK, a non-conformal mesh interface is an interface which does not have equal number of nodes on both sides, but it seems to me (visually) that I have an equal number of nodes on both sides (between zones):</p>
<p><a href="https://i.stack.imgur.com/qPY6B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qPY6B.png" alt="enter image description here" /></a></p>
<p>So how can I make sure that I have conformal mesh interface between the fluid zones and the porous zone?</p>
| |cfd|porous-medium|fluent| | <p>I found the issue addressed in Fluent user guide:</p>
<blockquote>
<p>Parts are groups or collections of bodies. Parts can include multiple
bodies and are then referred to as multibody parts. <em><strong>If your
geometry contains multiple parts then each part will be meshed with
separate meshes with no connection between them, even if they
apparently share faces.</strong></em></p>
<p>You can convert a geometry which has multiple parts into one with a
single part by using the Form New Part functionality in the ANSYS
DesignModeler application. Simply select all of the bodies and then
select Tools > Form New Part. If you have an external geometry file
that has multiple parts that you wish to mesh with one mesh, then you
will have to import it into the DesignModeler application first and
perform this operation, rather than importing it directly into the
ANSYS Meshing application.</p>
</blockquote>
<p>From DesignModeler:</p>
<p><a href="https://i.stack.imgur.com/9Xkg9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9Xkg9.png" alt="enter image description here" /></a></p>
<p>The whole geometry is now a one part (still with separate zones) with a conformal interface between the zones.</p>
| 8432 | How to force creating a conformal mesh interface between zones in Fluent? |
2016-04-15T16:09:16.783 | <p>As from the derivation we are able to see that the implicit density based solver (such as BTCS) is unconditionally stable for any courant number. So what is it, which limits the upper boundary of CFL while using it in practical applications?</p>
<p>I need this for simulations in ANSYS Fluent.
I have been told that non-linear effects start coming in, but I don't know what exactly are the non-linearities?</p>
| |cfd|ansys| | <p>Analytical stability results for backward-time, centered space (BTCS) discretizations only apply to the first order (linear) hyperbolic partial differential equation. The only reason why they are applicable to the navier stokes equation is because navier stokes tends to exhibit behaviors similar to this first order linear wave equation in some cases (e.g. slow moving laminar flow). The stability analysis is much easier to perform on linear equations such as the 1st order wave equation and provides somewhat of an estimate of timestepping stability in the non-linear hyperbolic PDE case. </p>
<p>The navier stokes equations have nonlinearities in the momentum equation (i.e. the nonlinear advection term $\nabla\cdot(\vec{u}\otimes\vec{u})$). As this non-linear term becomes stronger, the stability estimate based on BTCS discretization (for the 1st order linear wave equation) becomes less and less precise. Unfortunately, non-linear equations are extremely difficult (impossible?) to analytically derive stability conditions for different timestepping schemes. In practice, you will observe some deviations from the stability conditions for BTCS for the first order wave equation. Explicit timestepping schemes often need to be much smaller than 1 in order to observe convergence, while backward timestepping will have some limit to how large the time stepsize can be. Usually, this limit is larger than that which an explicit timestepping scheme would allow.</p>
<p>Heuristics on stability criterion for the navier stokes equations are generally obtained empirically via trial-and-error and are highly problem dependent.</p>
| 8439 | What limits my Courant number in Implicit Density based Solver? |
2016-04-16T17:24:27.867 | <p>I am curious to know the difference between a lead compensator (lead filter) and a PD controller with a low pass filter. From the theory I can see that the transfer function of both these controllers are same. By choosing appropriate gains and placing zeros and poles at right frequencies, we can obtain same bodeplot (magnitude and phase plot). So, what is the real difference between the two? Why use different names for the same controllers? </p>
| |control-engineering|transfer-function|feedback-loop|pid-control| | <p>A lead filter implies that the zero has a lower frequency than the pole. While a PD controller with a low-pass-filter does not necessarily imply that order. Also a lead filter (usually) does not have the zero and pole to far apart from each other, meaning that the bode diagram does not get very close to the asymptotes of +1 slope for magnitude and 90° for phase between the zero and pole. While the PD and lowpass in general can be spaced very far apart.</p>
<p>But you are right that there is theoretically not really a difference, besides that the PD and lowpass has an additional degree of freedom, due to the P action. Namely for both you have to place a zero and a pole.</p>
<p>However when implementing both you might get different results due to the way you might implement each filter when discretizing them. But usually there is only a significant difference when you get close to your Nyquist frequency.</p>
| 8448 | Is lead filter same as PD combined with a low pass filter? |
2016-04-16T20:09:47.670 | <p>I was very surprised when i found out that Newton's solution to the problem of solid of minimal resistance was a body with somewhat strange nose cone - the end of the nose cone is a plane perpendicular to the body's axis. So my question is, was Newton's calculation confirmed and used somehow by space programs?</p>
| |fluid-mechanics| | <p><strong>TL;DR</strong></p>
<p>Newton got some things spectacularly right, and other things spectacularly wrong. In this case, he didn't remotely understand the <em>real</em> physics of the problem - and neither did anybody else until a few hundred years after his death. </p>
<p>You shouldn't believe everything in Newton's writings on alchemy either - nor his predictions of the date of the end of the world based on his analysis of the Bible.</p>
<p><strong>Longer answer</strong></p>
<p>The great importance of Newton's work attacking this type of problem in <em>Principia</em> was not that he got almost all the details wrong, but that <strong>he attempted to solve the problems at all</strong> using mathematical modelling. </p>
<p>In Newton's time, there was no clear understanding of the difference between energy and momentum. The greatest achievement of <em>Principia</em> was in pinning down the concept of "momentum", and the relationship between "force" and "rate of change of momentum", as given by Newton's second law. And the mathematical edifice that Newton built on that foundation was of course spectacularly right.</p>
<p>But with no real concept of energy, Newton's ventures into fluid mechanics were much less successful. I very much doubt whether anyone working in the space program would be naïve enough to consider use Newton's models - and most likely they would know nothing about them.</p>
<p>To show the sort of errors in Newton's fluid dynamics models, it might be instructive to start with a simpler problem which he also considered in <em>Principia</em>, namely to calculate the speed of sound in air. Newton attempts to solve this by comparing the vibrations of the air with the vibrations of a pendulum (which he could both analyze and measure experimentally), which was a good way to start. But he had no notion of the concept of the internal energy stored in the fluid via the motion of individual molecules, which is related to the temperature of the fluid. In fact, in Newtons' time temperature measurement had barely progressed beyond the subjective and qualitative human notions of "hot" and "cold."</p>
<p>Using modern terminology, he incorrectly assumed that all the internal energy in the fluid was accounted for by changes in its pressure and volume. This led him to the equation of state $PV = \text{constant}$, instead of the correct adiabatic relationship $PV^{\gamma} = \text{constant}$ where $\gamma \approx 1.4$ for air. </p>
<p>After deriving the properties of air from experiment, this error produced a calculated value for the speed of sound that was about 10% slower than what had been measured. (His measured speed of 339 m/s was probably accurate, but there is no record of the temperature, humidity, wind speed, etc, when the measurement was made).</p>
<p>To account for this discrepancy, Newton makes the wild assumption (<em>Principia</em> Vol 2 Section 8 proposition 50) that if the particles making up the air are rigid spheres of finite size, and separated from each other by 10 times their diameter, then (by analogy with a "Newton's cradle" device with gaps in between the balls) the speed of sound would be about 10% higher than his calculated result. (It is perhaps unkind to wonder if he had forgotten his grand declaration "Hypotheses non fingo" - "I don't make speculative assumptions!") He doesn't attempt any mathematical justification of that idea, nor does he explain how the air particles align themselves with the direction in which the sound is travelling!</p>
<p>He then goes on to consider the effect of water vapor on the speed of sound. He seems to consider the particles of water vapor to be simply inert bodies that occupy some proportion of the volume of the air, and therefore that water vapor increases the speed of sound further by adding more "balls" to his supposed "cradle". Of course this logic is actually backwards, but Newton had no idea that the molecular weight of water is about half that of air, and the presence of water thus reduces the density of the air while having little effect on its compressibility.</p>
<p>His analysis of the air resistance of a projectile (<em>Principia</em> Book 2, Section 7, Proposition 34, theorem 28) is equally far from the truth. He attempts to do this by considering the change of momentum of the projectile as it collides with stationary particles of air and knocks them aside. He assumes the collisions are perfectly elastic, and therefore (in modern terminology) no mechanical energy would be converted into heat. That means he is assuming the fluid is inviscid. But the correct mathematical analysis of inviscid irrotational flow gives the result that the drag force is <strong>zero</strong> for <strong>any</strong> shape of body! Newton made a second mistake, in ignoring what happens to the air <em>behind</em> the projectile.</p>
<p>In fact, Newton's model is not too unrealistic for a completely different scenario, namely piercing a hole through a (fairly loose) granular material like earth. Indeed, the shape of the nose of a typical "mole" tool used for trenchless tunneling is the simple pointed cone shape that Newton claims would give the least drag, as shown here: <a href="http://en.terra-eu.eu/underground-piercing-tools/2-16-produkte" rel="nofollow">http://en.terra-eu.eu/underground-piercing-tools/2-16-produkte</a>. But any similarity between that scenario and the analysis of a hypersonic space capsule is purely conincidental.</p>
| 8450 | Was Isaac Newton's solution to the problem of solid of least resistance comfirmed in space programs? |
2016-04-18T03:04:12.720 | <p>I very much ruined 3 $10 diamond hole saws making this ceramic insulator out of a firebrick and Ryobi drill press. Drilling out the large core took 1-2 hours and a lot of arm strength.</p>
<p>Therefore, I will turn to custom ceramic shops or my own castable refractory molds. I am concerned that the custom ceramic shops will be too expensive though.</p>
<p>So what is a refractory cement available to me that can withstand the melting point of titanium? I.e. there is a Tungsten hotnozzle about 6mm x 10mm that goes at the bottom of this ceramic tube. It's for a 3D printer that prints metal (hopefully!). So, I don't need the entire piece to withstand that temp, but just at the end point and radiating around that.</p>
<p>However, when I search for castable refractory cement, it seems like only one product type comes up. </p>
<p><a href="https://i.stack.imgur.com/4UurB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4UurBt.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/ULAgo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ULAgot.jpg" alt="enter image description here"></a></p>
| |metallurgy|casting|ceramics| | <p>There are a number of approaches to this, depending on exactly what you want to achieve. </p>
<p>There are <a href="http://www.castreekilns.co.uk/dense-castable-189-c.asp" rel="nofollow">dense castable refractories</a> which will operate up to 1800C. For small quantities the best source is likely to be a ceramic kiln supplier. Industrial foundry supplier may have a wider range of products but tend to deal in larger quantities. </p>
<p>Another possible alternative is graphite, this needs to be shielded from oxygen to prevent oxidation, but then again so does molten titanium so you would need to do that anyway as titanium is very prone to porosity and embrittlement due to absorbing atmospheric gasses, even at temperatures well below its melting point. </p>
<p>Zirconia based refractories can also work well at very high temperatures. One option is to use a sofer insulating refractory, or indeed rigidised ceramic wool which is coated with a layer of zirconium slurry, this is the principal behind ceramic shell investment casting. </p>
<p>It is also possible that any shielding gas used will also provide a significant cooling effect and as such you may find that the case temperature is significantly less than the molten metal temperature, especially as pressurised argon will give you a significant cooling effect for free. </p>
<p>As an aside I'm not sure how much work you have done on the whole concept of extruding molten titanium through a nozzle but it seems a bit iffy to me as I am not aware of any current process which does anything remotely like this and for absolute certain in TIG welding any contact between the tungsten electrode and the base metal is normally considered a defect as the base/filler will instantly alloy with the tungsten and contaminate it. </p>
| 8462 | What is the highest-temp castable refractory cement available? |
2016-04-18T17:02:57.040 | <p>I have some questions in the basics of machinery terminology </p>
<ol>
<li><p>Can a spring ever be a link? Most books (eg. Ham et al. 58, Shigley pg. 5) consider links to be rigid. However, I have come across articles using springs as links.</p></li>
<li><p>Are joints always the same thing as kinematic pairs, or do they ever differ in any particular circumstances?</p></li>
<li><p>Is mechanism a subset of kinematic chain or is chain a subset of mechanism? Some books mention if there is no fixed link, then the combination cannot be called a mechanism but can be a kinematic chain.</p></li>
</ol>
| |mechanical-engineering|terminology|kinematics| | <p>@katipra I am adding this as another answer since it was too long for the comment section. But it's a reply to your aforementioned comment.</p>
<p>Good news. Based on the actual definitions of Merriam-Webster of machine vs. machinery:</p>
<p>"Machine: 1) an assemblage of parts that transmit forces, motion, and energy one to another in a predetermined manner.
2) an instrument (as a lever) designed to transmit or modify the application of power, force, or motion.
3) a mechanically, electrically, or electronically operated device for performing a task."</p>
<p>"Mechanism: 1) a process, technique, or system for achieving a result
2) mechanical operation or action."</p>
<p>Therefore, the best conclusion would be that machine would be the most suitable word for an IC engine.</p>
<p>Also, I have found a reliable link in NASA's database that mentions "When discussing engines, we must consider both the mechanical operation of the machine and the thermodynamic processes that enable the machine to produce useful work."</p>
<p>[<a href="https://www.grc.nasa.gov/WWW/K-12/airplane/icengine.html]" rel="nofollow noreferrer">https://www.grc.nasa.gov/WWW/K-12/airplane/icengine.html]</a></p>
| 8473 | Questions on machinery terminology |
2016-04-19T01:02:51.577 | <p>What kinds of sensors can be used to monitor the condition of a hard disk?</p>
<p>We want to be able to detect when a hard disk is not operating correctly.</p>
<p>Unexpected vibrations may be caused by spindle problems (probably low frequency), or by head crashing (high frequency).</p>
| |mechanical-engineering|electrical-engineering|motors|sensors| | <p>There are two types of sensors commonly used in hard disks to detect vibration and head crashing. They are:</p>
<ol>
<li><strong>Accelerometer</strong>: Accelerometers are used to detect sudden vibrations. </li>
<li><strong>Sudden Motion Sensors:</strong> These are mostly used in Apple Notebooks to detect a sudden drop of a laptop and to disengage the hard drive. SMS system parks the hard drive and restart the hard drive when stability is detected. SMS include a triaxial accelerometers. </li>
</ol>
<p><strong>References:</strong></p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Active_hard-drive_protection" rel="nofollow">Active hard-drive protection</a></li>
<li><a href="https://en.wikipedia.org/wiki/Sudden_Motion_Sensor" rel="nofollow">Sudden Motion Sensor</a></li>
<li><a href="http://Mac%20notebooks:%20About%20the%20Sudden%20Motion%20Sensor" rel="nofollow">Mac notebooks: About the Sudden Motion Sensor</a></li>
<li><a href="http://www.analog.com/library/analogDialogue/archives/39-11/hdd.html" rel="nofollow">Using Dual-Axis Accelerometers to Protect Hard Disk Drives</a></li>
<li><a href="http://www.eetimes.com/document.asp?doc_id=1271337" rel="nofollow">Accelerometers protect hard drives</a></li>
</ul>
| 8477 | What sensors should be used to analyze vibration of a hard disk? |
2016-04-19T18:56:22.023 | <p>In a tensile test the peak load was 69kN, but the point of failure load was 60kN. Why is it lower?</p>
| |mechanical-engineering|structural-engineering|civil-engineering|stresses| | <p>That is due to <a href="https://en.wikipedia.org/wiki/Necking_(engineering)">necking</a>.</p>
<p>When a bar is under tension, two opposing mechanisms take place:</p>
<ul>
<li>The bar loses cross-sectional area since it attempts to retain its volume while being stretched (see <a href="https://en.wikipedia.org/wiki/Poisson%27s_ratio">Poisson's ratio</a>).</li>
<li>Once the bar begins to suffer plastic deformations, it undergoes <a href="https://en.wikipedia.org/wiki/Work_hardening">stress hardening</a> and it's elastic modulus rises.</li>
</ul>
<p>So long as the increase in elastic modulus is greater than the loss of area, the stress (or force) measured will increase. At a certain point, however, the material starts to lose area faster than it hardens, which means that the force required to continue deforming the element starts to diminish, until the element reaches its ultimate deformation capacity and snaps.</p>
<p>Now, if you'd been measuring the force and dividing by the element's original cross-sectional area to get stress results, then these would have the same profile as the force results: increasing up to the peak load and then dropping to point of failure. This is known as <strong><em>engineering stress</em></strong>.</p>
<p>If, however, you use special equipment to measure the actual cross-sectional area as it changes during the test, then you'll see that the stress actually increased during the entire test due to stress hardening. This is known as <strong><em>true stress</em></strong>.</p>
<p>Here's a graph showing the difference between engineering and true stress. The red line represents engineering stress and the blue, true stress.</p>
<p><a href="https://i.stack.imgur.com/PPrjr.png"><img src="https://i.stack.imgur.com/PPrjr.png" alt="enter image description here"></a><br>
<sub>Source: <a href="https://en.wikipedia.org/wiki/Hooke%27s_law#/media/File:Stress_v_strain_A36_2.svg">Wikipedia</a></sub></p>
<p>As the name implies, engineers tend to use engineering stress for the simple reason that it is infinitely easier to measure and design with. And as the diagram above demonstrates, the difference between engineering and true stress in the elastic range where we spend 95% of our time is very small.</p>
| 8488 | Why is the point of failure load lower than the peak load? |
2016-04-19T19:44:19.940 | <p>I have a tube, at the bottom of the tube is burning wood and an air inlet. How can i increase the speed of the exhaust gas coming out of the tube? I thought about "bottle-necking" the exhaust pipe to make it go faster (Bernoulli's principle) but is there any other way?</p>
| |fluid-mechanics|thermodynamics|combustion| | <p>i am not sure about bottelnecking what does it means but the only natural way to increase the exhaust velocities is to place a <strong>NOZZLE</strong> at the outlet as it converts the pressure energy into kinetic energy(<strong>From Steady Flow Energy Equation)</strong> and to have more kinetic energy you should have high rate of combustion that will produce high pressure resulting into high velocities through a nozzle.
for the non natural processes you can use a centrifugal/reciprocating pump to achieve higher rate of discharge</p>
| 8489 | How to increase the speed of exhaust gas? |
2016-04-20T15:02:14.247 | <p>I am analyzing an industrial ventilation system that at some point operates two industrial size blowers in parallel. They often show oscillating behaviour: one blower briefly draws more power / generates more flow, and the other performs less. Then the roles reverse. The only practical, known way to reduce the oscillation is to throttle down the vane valve in front of the fan and then open the valve again. The blowers are almost (but not completely) identical.</p>
<p>I seems that both blowers affect each other in some unstable way. Is this a known phenomena? I'm not finding anything when searching for fan/blower/pump oscillation/resonance/hysteresis.</p>
| |mechanical-engineering|fluid-mechanics|pumps| | <p>The reason is due to slight instability of the fans, and bad fan stability selection. To begin with, all backwards inclined fans have, to some extent, an unstable range. All fans are designed to operate at an RPM using an RPM chart. This includes the stable and unstable ranges of the fan:</p>
<p><a href="https://i.stack.imgur.com/Ax5cF.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ax5cF.gif" alt="enter image description here"></a></p>
<p>So long as the fan is in the "downward slope", this is considered acceptable. This has been discussed as <strong><em>fan surge</em></strong> with <a href="https://engineering.stackexchange.com/questions/14604/how-to-design-a-vacuum-climbing-robot/14633#14633">other answers</a> on this site. One of the big notices with fan surge is that it stops at the local maximum of the fan, and goes no farther. Many times, a single fan is considered to operate at this point and this is fine for operation. However, operating multiple fans in parallel can cause problems if they are in the normally stable region, but above the maximum pressure at no flow point:</p>
<p><a href="https://i.stack.imgur.com/Xki3Y.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xki3Y.gif" alt="enter image description here"></a></p>
<p>This is because the double fans can oscillate as you describe, with one fan operating at the left point of the curve, and one fan operating at the right point of the curve, then rapidly switching back and forth. Only by ensuring that the static pressure across the fans is less than the no-flow pressure, can you ensure completely stable operation.</p>
<p>For more research, see below some additional sources. <a href="https://www.ksb.com/centrifugal-pump-lexicon/parallel-operation/191648/" rel="nofollow noreferrer">Math Source</a> <a href="http://www.greenheck.com/library/articles/42" rel="nofollow noreferrer">Graphical Source</a></p>
| 8505 | Why do parallel fans/blowers show an oscillatory behavior? |
2016-04-20T15:59:05.797 | <p>We want to monitor hard disks for present problems or pending failure.</p>
<p>Can the Fast Fourier Transform (FFT) of a given hard disk in question, then, be compared with a reference FFT for a perfectly working one, and then conclude hard disk may be faulty or about to fail if its FFT differs significantly from the reference?</p>
<p>FFT would be used to detect unexpected frequencies, or harmonics, that do not normally occur in a 'healthy' hard disk. FFT would input vibrations from hard disk, and output the collection of vibration's constituent frequencies. This would be compared to the frequency domain function of 'healthy' hard disk.</p>
| |mechanical-engineering|electrical-engineering|motors|sensors| | <p>Im pretty confident in that you can see some changes in the fourier analysis if you have a sensitive enough sensor in the right place and the failure is mechanical. Similar methods are used to measure the health of bearings with success.</p>
<p>However, you do not know what you are looking for. Unless your a very skilled simulation guy or can find some analysis which tells you what your looking for then you need some empirical data. Go forth and measure a bunch of drives and try to see how they differ. Get hold of ones that have failed and see what the difference is. Then try to predict which drive is going to brake and monitor that.</p>
| 8508 | Can Fast Fourier Transform be used to monitor for fault vibrations in a hard disk? |
2016-04-22T10:23:38.033 | <p>I'm trying to figure out what kind of pump I should use to boil water at low temperature. Here is my idea of the setup:</p>
<p><a href="https://i.stack.imgur.com/cQb6C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cQb6C.png" alt="enter image description here"></a></p>
<p>The dark blue water comes in, the pump sucks down the light-blue water. This will reduce the pressure in the top, forcing the dark blue water start boiling.</p>
<p>The water then condenses and gets pumped out.</p>
<p>An important condition is of course the temperature of the incoming water. This can be calculated using the <a href="https://en.wikipedia.org/wiki/Antoine_equation" rel="nofollow noreferrer">Antoine equation</a>, </p>
<p>$$p =10^{\left(8.07131 – \frac{1730.63}{233.426+T}\right)}$$</p>
<p>where $p$ is pressure in mmHG, and $T$ is temperature in degrees Celsius.</p>
<p>I'm trying to figure out what how I can translate the boiling of water to psi. I get confused because as some water gets pumped away, that will be replaced by newly condensed water, thus increasing the pressure again.</p>
<p>I'm assuming I should use a positive-displacement pump. Please let me know if this is a bad idea for some reason.</p>
| |mechanical-engineering|fluid-mechanics|thermodynamics| | <p>You can use a water pump to evaporate water. But you shouldn't.</p>
<p>The evaporation chamber needs to be closed, except for the suction pipe. Otherwise when you lower pressure water (or air) will flow in, raising pressure again. So you can create small batches of vapor. </p>
<p>The water will evaporate where pressure is lowest, this is likely to be your pump inlet. <a href="https://en.wikipedia.org/wiki/Cavitation#Pumps_and_propellers" rel="nofollow">Not what you want</a>. You need to make sure that the ambient pressure in the evaporation chamber is lower than vapor pressure at whatever temeprature your water has. At the same time, ambient pressure + hydraulic head between level in chamber and pump inlet needs to be higher <a href="https://en.wikipedia.org/wiki/NPSH" rel="nofollow">by a margin</a>, else you have evaporation in your pump. Not good.</p>
<p>Then, you have low pressure, low temperature vapor. What do you want to do with it and how do you transport it? If you use a blower, the same blower could also create the vacuum for your evaporation and save you all the pump headaches.</p>
<p>So you could create vapor with a water pump in a batch process, but I fail to see an aplication where this makes sense.</p>
| 8534 | Can I use a water-pump to lower pressure and boil water? |
2016-04-22T14:25:04.783 | <p><a href="https://i.stack.imgur.com/PTClt.png" rel="noreferrer"><img src="https://i.stack.imgur.com/PTClt.png" alt="enter image description here"></a></p>
<p>Hello, I've been asked to show that:
$$d\Gamma=\tau 2\pi r^2 dr$$</p>
<p>Where $\Gamma$ is the torque applied to the shaft and $\tau$ is the shear stress of the sample fluid at radius $dr$. So, I know that $d\Gamma= Force\cdot dr$ and that $Force = \tau \cdot Area$ and this is where I start encountering issues. If we're looking at some radius $dr$, isn't the area $2\pi dr $ or just the circumference of the local circle? So I end up with $d\Gamma = \tau 2 \pi dr\cdot dr$ which isn't right, can someone help?</p>
| |mechanical-engineering|fluid-mechanics|applied-mechanics| | <p>Your Average MechEng's answer is fantastic and it is often how I explained the area of a thin circular shell to my students. But the mathematical derivation of such formula might be of interest to some. Using the above answer's diagram:</p>
<p><a href="https://i.stack.imgur.com/X3XXO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X3XXO.jpg" alt="enter image description here"></a></p>
<p>we are interested in computing the area between the two dashed lines. The inner circle has area $\pi r^2$ while the outer circle has area $\pi (r+\Delta r)^2$. The area between the two is obtained by subtraction:
$$
\begin{align}
\text{Area} &= \pi (r+\Delta r)^2 - \pi r^2 \\
&= \pi (r^2 + 2r\Delta r + \Delta r^2 - r^2) \\
&= \pi (2r\Delta r + \Delta r^2)
\end{align}
$$
Taking the limit as $\Delta r$ goes to zero yields the desired answer.</p>
<p>As an aside, it does not seem like any of these two answers resolve your issue, although they get you closer.</p>
| 8541 | Force Balance Question |
2016-04-22T18:47:45.563 | <p>An uncold-worked brass specimen of average grain size 0.009 mm is heated to 600 degrees Celsius for 1000s, what is the average grain size based on this graph?</p>
<p><a href="https://i.stack.imgur.com/pRS7L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pRS7L.png" alt="enter image description here"></a></p>
<p>How to tackle this problem? I don't know if I am using the graph correctly.
My approach was:</p>
<p>1000s is 16.67 min, $\log(16.67) = 1.22$</p>
<p>So $\frac{22}{100}$ between 10 and 10<sup>2</sup>. Which is about 0.67 cm to the right of 10 on x-axis.</p>
<p>Now lets estimate this corresponds to 10<sup>-1.333</sup> on the y-axis of the 600 degrees celcius graph line. So 0.046 mm.</p>
<p>But this does not seem correct, because I did not incorporate the 0.009 mm starting position. Which corresponds to heat treatment time of about 1 min at 600 degrees so should I add 1 min to the 1000s? And thus look look at the point on the graph at 17,76 min?</p>
<p>The answer in book seems way off, they talk about average grain size 0.2 mm. To me that seems like they looked at the graph of 700 degrees Celsius.</p>
| |materials|metallurgy| | <p>Grain growth is the movement of grain boundaries by diffusion to reduce the grain boundary area. This graph tells that at 600 degrees, a treatment of one minute shall give the y coordinate of the 600 degrees graph. So treating a sample for 1000s shall mean heating the sample at 600 from time t=0.0. The starting point is inherent in the process itself. </p>
| 8547 | Estimate the grain size after grain growth |
2016-04-23T13:21:24.327 | <p>With gravity camera stabilisers like below, how come counterweights are needed to stabilise the load (camera)?</p>
<p><a href="https://i.stack.imgur.com/CGmS7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CGmS7.jpg" alt="enter image description here"></a></p>
<p>Is it to ensure that the handle joins at the centre of gravity? If so, why does it have to be at the centre of gravity?</p>
| |mechanical-engineering| | <p>A camera system is more sensitive to rotational shake than any translation. That is because translation causes a shift proportional to the noise but rotation causes an error proportional to the distance to the target.</p>
<p><img src="https://i.stack.imgur.com/GqjQ2.png" alt="Camnoise"></p>
<p><strong>Image 1</strong>: Example of noise in a plane, note that in 3 dimensions there are more rotations.</p>
<p>Conservation of momentum, means that more mass is slower to move around. When it comes to rotation the distribution of this mass affects the way something resists rotation. We call this moment of inertia.</p>
<p>Inertia brings "slowness" to the system. Net effect is that the harder it is to rotate, the more force you need for the rotation to happen. Thus the system rotates less from the noise.</p>
<p>By putting the weight far away from the joint you make the moment of inertia bigger. Meaning its harder to rotate (around the axis that has furthest distance to weight) thus damping the system more. At the same time the overall weight of the system does not need to grow as much. </p>
<p>For similar reasons, you want the center of gravity be below the ball joint so that the camera stays upright. Or more exactly, the system will orient so that the center of gravity is under the ball joint, so you want to design it so that this benefits you. Also you want to avoid having a lot of weight around the ball joint on a horizontal plane so that turning the camera horizontally is not too heavy. This way, the system prefers motions that are more natural for your camera work.</p>
| 8550 | How come counterweight is needed to stabilise camera? |
2016-04-23T16:34:08.953 | <p><a href="https://i.stack.imgur.com/dzdEC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dzdEC.png" alt="enter image description here"></a></p>
<p>The flux in the gas can be written as:</p>
<p>$$N_{1}=k_{p}(p_{1}-p_{1i})$$</p>
<p>And the flux in the liquid can be written as:
$$N_{1}=k_{x}(x_{1i}-x_{1})$$</p>
<p>And we know that $$x^*_{1}=\frac{p_{1}}{H}$$</p>
<p>How can I use this to find the overall mass transfer coefficient?
I know that the driving force will just be $(p_{1}-Hx_{1})$ but what's $K_{g}$ and more importantly, how is it derived?</p>
<p>EDIT: I'll give the best answer to anyone who can even give a hint</p>
| |heat-transfer|chemical-engineering|process-engineering|diffusion| | <p>Convert both the pressure and concentration terms as mole fractions.
Now you have y for gas phase mole fraction and x for liquid phase mole fraction.
Henry's law is taken as am equilibrium relation in your question, so the equilibrium diagram(y vs x) is just a straight line through the origin with slope H'.
Taking the gas side mass transfer coefficient on mole fraction basis as ky.
So, on the equilibrium diagram, the bulk point is given as M(x0,y0) and interfacial concentration as N(xi,yi).<strong>[Note: N must be on the equilibrium curve as the interface concentrations are always in equilibrium.]</strong>
Now, we take a quantity (yb*) that is defined as the mole fraction of solute in gas phase that can remain in equilibrium with (xb), the mole fraction in bulk liquid phase.</p>
<p>N = ky(yb-yi) = kx(xi-xb) = Kg(yb-yb*)</p>
<p>(yb-yb*) = (yb-yi) + ((yi-yb*)/(xi-xb))*(xi-xb)</p>
<p>N/Kg = N/ky = H'*N/kx <em>(since yb</em> is in equilibrium with xb, by definition)*</p>
<p>This give the overall gas phase coefficient Kg = (1/ky) + (H'/kx).
Overall liquid phase coefficient can be determined similarly.</p>
<p>NOTE : The equilibrium diagram described above is used when the equilibrium relationship isn't linear.</p>
| 8552 | Derivation of Overall Mass Transfer Coefficient |
2016-04-24T15:09:23.040 | <p>I read in a documentation to some adhezive: <em>the tensile strength is 13 N/mm<sup>2</sup></em>. How many kilograms can the 10x10mm bonding hold?</p>
<p>I tried to calculate it by the following:</p>
<p>$$\begin{align}
F &= m/g \\
\therefore m &= Fg
\end{align}$$</p>
<p>the strength is in N/mm<sup>2</sup> e.g. that is</p>
<p>$$\begin{align}
p = F/A \\
\therefore F = pA
\end{align}$$</p>
<p>substituting the $F$ in $m = Fg$, I got</p>
<p>$$m = pAg$$</p>
<p>Using real numbers,
- tensile strength = 13 N/mm<sup>2</sup> = 13*10e6 N/m<sup>2</sup>
- A = 10x10mm (100mm<sup>2</sup> = 0.0001m<sup>2</sup>)</p>
<p>$$m = 13*10e6 * 0.0001 * 9.81 = 12753 \text{ kg} $$</p>
<p>The result doesn't look right - the 10x10mm bonding imho can't hold 13.000kg weight.</p>
<p>Where did I make a mistake? So the question is: how to convert the 13 N/mm<sup>2</sup> tensile strength to kilograms (e.g. how many kilos I can hang to the bonded 10x10mm plates?)</p>
| |structural-engineering|stresses| | <p>Pay attention to your unit analysis. You multiplied by the acceleration due to gravity, rather than dividing by it. Unit analysis would have shown this to be nonsense.</p>
<p>$$13 \frac{N}{mm^2} \cdot 100 mm^2 = 1300 N = 1300 \frac{kg \cdot m}{s^2}$$</p>
<p>Therefore, to get kg, you need to divide by an acceleration:</p>
<p>$$\frac{1300 \frac{kg \cdot m}{s^2}}{9.81 \frac{m}{s^2}} = 132.5 kg$$</p>
<hr>
<p>EDIT: In terms of the symbolic analysis, your very first statement was wrong: Force equals mass <em>times</em> acceleration, so you should have written it as</p>
<p>$$F = m\cdot g \text{ and } m = \frac{F}{g}$$</p>
<p>Then, when you combine that with the equation for pressure, you get</p>
<p>$$m = \frac{pA}{g}$$</p>
| 8560 | Tensile strength to weight conversion |
2016-04-24T19:21:57.257 | <p>The precast concrete stairs at the home I recently purchased do not have a footing. Lack of a footing coupled with drainage issues has caused them to sink over time. I'm going to raise them and get them properly supported underneath.</p>
<p>To do that I want to clamp them near their center of gravity (along the length), so that I can jack them up back to where they belong and minimize the risk of breaking the concrete (I'll continuously add temporary support underneath them as I slowly jack them up).</p>
<p>I have dug out the bottom of the steps and determined that they are monolithically poured and solid.</p>
<p>What is the measurement along the length of the steps where the center of gravity will be?</p>
<p><a href="https://i.stack.imgur.com/Zcsx6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zcsx6.png" alt="enter image description here"></a></p>
<p>Edit: This is not homework. Here are the stairs in question:</p>
<p><a href="https://i.stack.imgur.com/AsSPU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AsSPU.jpg" alt="enter image description here"></a></p>
<p>You can see how they are pulling away from the foundation and have rotated forward and dropped. The sidewalk slab in front of it has also sunk and needs to be raised, so it is worse than it seems in this photo.</p>
<p>I attempted to solve this by hand because it seems trivial. I'm glad I asked, because the solution I came up with was wrong. I assumed I could reduce it to working with the ratio of cross sections which are simply three rectangles (since they all have the same width).</p>
<p>If I take the area of each step individually I get:</p>
<p>$$\begin{align}
17.5 \times 18 &= 315 \\
14 \times 11 &= 154 \\
47.5 \times 3 &= 142.5
\end{align}$$</p>
<p>I noticed that if I add the bottom two steps together, they are a smaller combined area (and thus volume) than the large step [pretty sure this is where I went wrong ... when you multiply by width this is no longer the case]. So it appeared to me at the time that the COG would be somewhere in the large step. I'll denote the COG as $c$.</p>
<p>So I then used this to give me $c$:</p>
<p>$$18 \times c = 18 \times (17.5 - c) + 154 + 142.5$$</p>
<p>That worked out to 16.98 (17 inches for all practical purposes).</p>
<p>However, after seeing @Chris Johns solution, I realize that what I had in my brain is not correct, and anybody with sinking stairs should ignore everything above.</p>
| |mechanical-engineering|structural-engineering| | <p>The equation to find the center of gravity of an object which can be subdivided into smaller parts is the following:</p>
<p>$$\text{CG} = \dfrac{\sum \overline{y}_iA_i\rho_i}{\sum A_i\rho_i}$$</p>
<p>where $\overline{y}_i$, $A_i$ and $\rho_i$ are the <a href="https://en.wikipedia.org/wiki/Centroid">centroid</a>, area and specific weight of each part, respectively.</p>
<p>In your case, the specific weight of all parts is equal, and so the equation can be simplified to</p>
<p>$$\text{CG} = \overline{y} = \dfrac{\sum \overline{y}_iA_i}{A}$$</p>
<p>Therefore, in such a case, the center of gravity of the object is equal to its centroid ($\overline{y}$). Also, to be clear, the $A$ in the denominator is the object's total area.</p>
<p>So, in this example, the solution is:</p>
<p>$$\begin{align}
\overline{x}_1A_1 &= \left(\dfrac{17.5}{2}\right) \times \left(18 \times 17.5\right) \\
\overline{x}_2A_2 &= \left(17.5 + \dfrac{14}{2}\right) \times \left(10.5 \times 14\right) \\
\overline{x}_3A_3 &= \left(17.5 + 14 + \dfrac{47.5}{2}\right) \times \left(3.0 \times 47.5\right) \\
A &= \left(18 \times 17.5\right) + \left(10.5 \times 14\right) + \left(3.0 \times 47.5\right) \\
\color{red}{\overline{x}} &= \color{red}{\dfrac{\sum \overline{x}_iA_i}{A} = 23.5} \\
\overline{y}_1A_1 &= \left(\dfrac{18}{2}\right) \times \left(18 \times 17.5\right) \\
\overline{y}_2A_2 &= \left(\dfrac{10.5}{2}\right) \times \left(10.5 \times 14\right) \\
\overline{y}_3A_3 &= \left(\dfrac{3.0}{2}\right) \times \left(3.0 \times 47.5\right) \\
\color{red}{\overline{y}} &= \color{red}{\dfrac{\sum \overline{y}_iA_i}{A} = 6.3}
\end{align}$$</p>
<p>Your center of gravity is 23.5" from the left-hand side of your image (from the exterior face of the upper step) and 6.3" from the bottom of your image.</p>
<p>That being said, don't think that just because you raise the steps from their center of gravity that they won't get damaged. If you just do that, then the steps may become like a dual-cantilever, in which case your thin "step" (if you can call it that) will risk breaking apart due to its low height and therefore low moment of inertia. Now, if what you actually want is help on raising your steps without damaging them, I suggest you ask another question referencing this one.</p>
| 8564 | Center of Gravity of Precast Concrete Steps |
2016-04-27T06:04:40.573 | <p>How is the position of a train modeled in a computer control system?</p>
<p>I can think of two basic approaches but I am not sure which (if any) are used in the real world.</p>
<h1>Lat, Long</h1>
<p>Ex: (-42, 23) </p>
<p>The position of the train is modeled by its real world position. The only issue I could see with this is extra work would have to be done at some stage to determine what track the train is on.</p>
<h1>Distance, Track</h1>
<p>Ex: (20m, Track 34-B) </p>
<p>The position of the train would be modeled by its distance from the start of the track, and the track it was on. This isn't exactly as intuitive right off the bat but I think it makes more sense.</p>
<p>I only ask because earlier I was reading some about PTC and as I was reading this popped into my head.</p>
<h2>Edit:</h2>
<p>As brought up by ratchet freak it most likely depends on the sensors on a track. </p>
<p>I was wondering if there is anyone out there who is or currently was part of the development of such a PTC system who could speak to which method is used. </p>
<p>Or if there is a link to a detailed specification of a PTC system (May exist because the conversion of railways is partially managed by the government). </p>
<p>My guess is probably something similar to (Distance, Track) for data gathered by on track sensors and then GPS is used as a fallback which can be converted into (Distance, Track).</p>
| |civil-engineering|software|rail|transportation| | <p>Speaking personally (as someone who deals with this sort of thing for a US Class I on a regular basis) -- operational and maintenance personnel find railroad locations the same way you and I find highway locations -- milepost + track name, and named locations called "stations" in US railroad parlance.</p>
<p>The former are not far off in usage from their highway counterparts -- albeit not always one mile apart due to changes in the track alignment. In areas where more than one track is present, the various tracks are named as well -- this may be something like "Track Two" for a mainline track in a multiple main territory or something more obscure like "50" or "Piglet Lead" for tracks in yards and industrial areas.</p>
<p>The latter may be akin to a named/signed highway junction, or simply a named point along the tracks marked by a sign. These are more commonly used by higher-level people, such as folks in Customer Service, as they don't care as much about what precisely the train is doing at a given station.</p>
<p>Communication Based Train Control (CBTC) systems may either use GPS lat/long and then convert it to a track name and milepost based on an internal database of the railroad's layout (this is what Positive Train Control or PTC does), or use beacons set in the track that act as "electronic mileposts" (the European Rail Traffic Management System/ERTMS approach, at least at Level 1).</p>
<p>Of course, along with this all, you need to know what route (subdivision) you are on to begin with -- that's akin to the highway number on a road. It doesn't do you any good at all to know you're at milepost 100.000000 on the main track if you don't have the foggiest clue what subdivision you're in!</p>
| 8610 | How to notate the position of a train? |
2016-04-27T13:56:56.740 | <p>I have acquired an old loading dock built in 1930 that has toe-bearing rafters in addition to undersized purlins. As you can imagine, many of the rafters have sagged. Also, because of the toe-bearing design of the rafter, the building has no eaves.</p>
<p>There are many articles about avoiding toe-bearing design. But there aren't many articles about remedying the situation in retrospect. Here is a drawing of the existing situation. The small white block in the middle of the rafter is a purlin.</p>
<p><a href="https://i.stack.imgur.com/XR0n7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XR0n7.jpg" alt="enter image description here"></a></p>
<p>So here are some questions and thoughts:</p>
<ol>
<li><p><strong>Inexpensive Fix</strong> - The articles state that you could use some type of joist hanger to possibly fix the problem, but it's not clear to me at all what type of joist hanger would help me. This would be a low cost way to fix the issue. I could:</p>
<ol>
<li>Jack up the sagging rafters</li>
<li>Sister up a new rafter, and </li>
<li>Use a joist hanger to help prevent cracks in the re-entrant corner.</li>
</ol></li>
</ol>
<p>Is there a Simpson Strong-tie (or some other brand) hanger that would help solve this issue?</p>
<ol start="2">
<li><p><strong>Medium cost Fix</strong> - I would really like eaves, so I was thinking another much more involved approach might be to:</p>
<ol>
<li>Support the existing rafter with a false wall,</li>
<li>Lower the wall by half a block, and then</li>
<li>Sister up a new rafter that has a proper birdsmouth cut that will allow the rafter to sit on the heel.</li>
</ol></li>
</ol>
<p>The sistered rafter could now extend beyond the wall, giving me an eave. Here's a picture of what I am thinking about:</p>
<p><a href="https://i.stack.imgur.com/NZzJa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NZzJa.jpg" alt="enter image description here"></a></p>
<p><strong>Edit</strong>: Ugh, it's not clear from the angle I posted, but the sistered rafter is sitting on it's heel on the lowered block/sill plate. It's not hanging in the air.</p>
<p>A properly engineered sistered rafter can take the load from the purlin on down, can't it?</p>
<p><strong>Edit:</strong> Based on a comment, here are more details: The existing rafters are 2x6, 24" centers. The pitch is 4/12. Just from visual inspection, the re-entrant corner on the birdsmouth appears to be right around half the thickness of the 2x6, so it seems like it's effectively a 2x3. I believe the purlin is about 9' from the birdsmouth, but I would need to remeasure that, if it's critical.</p>
<ol start="3">
<li><p><strong>Most expensive fix</strong> - Replace the entire roof with prefabbed trusses. Seems like overkill, but that sort of depends on the answers to the questions above.</p></li>
<li><p>Better/cheaper/easier ideas?</p></li>
</ol>
| |structural-engineering| | <p>OK, so after having a couple more roofing gurus come out, I just had one with an "ahha moment" that seems so obvious now looking back on it. So far this seems to be the best solution to my particular problem, and may help somebody else in the future with a toe-bearing rafter situation.</p>
<p>The idea is simply build a new roof on top of the old one. </p>
<p><a href="https://i.stack.imgur.com/HaBR7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HaBR7.jpg" alt="enter image description here"></a></p>
<p>By running new 2x6 rafters from the ridge down over the existing sheathing, I get the overhang I want, get my R-19 insulation, don't have to try and take the sag out of the old rafters, don't have to worry about an expensive demolition. I will plan on shimming the new rafters over the purlins where the old rafters have sagged, to make sure the new rafters are supported by the purlins from Day 1 and don't start sagging too.</p>
<p>Some questions that come to mind with this solution are:</p>
<ol>
<li>will the existing steel ridge beam handle the weight of the new roof and any snow loads?</li>
<li>how best to fasten the new rafters at the tail end?</li>
<li>can the new rafters remove some stress off the old rafters?</li>
</ol>
<p>Anyhow, thought I'd share this contractor's simple, yet clever solution.</p>
| 8613 | Fixing a Roof with Toe-bearing Rafters |
2016-04-27T18:13:24.710 | <p>I have assembled a small DIY drip-irrigation system for my terrace garden. Please have a look at the attached image. I switch on a small pump to start the drip-irrigation system and then switch it off. But even after that, the water keeps flowing through the system and stops only when I physically lift the pump out of water.</p>
<p>How can I stop this water flow without requiring any physical action?</p>
<p>Please note that I plan to automate the switch-on/off of the pump using a timer so that it functions without requiring my physical presence.</p>
<p><a href="https://i.stack.imgur.com/dcnMT.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/dcnMT.jpg" alt="Drip Irrigation System Setup"></a></p>
<p>========================================================================</p>
<p>I tried the suggestion of "Making a pinhole in the highest point of the pipe." It worked like a charm and solved the issue I was facing. Completely loved it, more so, because it does not involve procuring new stuff.</p>
<p>I also like the option of using a solenoid-valve, but have not tried it yet. Will use it when I install such a system again where the tank is at a considerably higher place than the plants.</p>
<p>Thank you everyone who took the effort to write an answer, and that too with detailed explanations.</p>
| |fluid-mechanics|siphon| | <p>I had this problem and solved it by looping the outlet pipe above the tank water level, cutting and fitting female hose connectors, and fitting them to a three-way male connector. On the third outlet I fitted a very short piece of hose with a standard connector, and at the other end of the pipe I used a Gardena 'water-stop hose connector'. In operation, pump pressure blocks the water-stop connector, but when the pump stops, air flows back through the water stop connector effectively breaking the siphon. The image was taken before it was tidied up. Cost is under NZD20, USD10.</p>
| 8619 | How to stop water flow in a siphon? |
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