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2017-05-01T12:17:38.737 | <p>When brazing HVAC linesets, the correct method is to flow an inert gas through the line in order to prevent interior oxidation of the copper and brazing products. However, as a specifier, I can say that this needs to be done until I'm blue in the face, but contractors (in my experience) are notorious for skipping this step. It adds cost, time, and a system where this is not done will not fail immediately, but it will contribute to the premature failure of the system. In the long term, this simply looks like a system that ran it's course, rather than the product of poor installation practices.
Is there a way to tell by inspection after installation that this practice was followed, or are you at the mercy of the contractor if someone is not watching?</p>
| |hvac|specifications| | <p>If the system did not have an inert gas to purge all of the oxygen out, the system would be very hard to reach the desired micron level with in the vacuum. </p>
| 15075 | Inert gas use during HVAC line installation |
2017-05-02T08:49:26.620 | <p>At what height do we splice the vertical bars in columns and shear walls is it at the middle of the element or at the bottom ( Slab level) and why ?</p>
| |structural-engineering|civil-engineering|design|structural-analysis|detailing| | <p>In theory, a splice behaves exactly like a single continuous bar, so there should be no effect as to where you actually do the splice.</p>
<p>However, in the real world it depends.</p>
<hr>
<p>If you're dealing with a compression-only column or wall, then I'd suggest splicing near (but not <strong><em>at</em></strong>) the slab level.</p>
<p>One reason is simple ease of construction. For example, if you must splice between floors 2 and 3 of a building and you do so just above floor 2, then the construction will look like this:</p>
<ul>
<li>Pouring of column from floor 1 to 2 without the splicing rebar (but with protruding rebar awaiting the splice).</li>
<li>Pouring of floor 2's slab</li>
<li>Working over floor 2's slab, the workers now have easy access to the splicing area, without needing to use scaffolding or anything of the sort to reach it.</li>
</ul>
<p>Obviously, the splicing shouldn't occur too close to the slab otherwise you might end up with a jumble of rebar: you have to fit twice the column reinforcement due to the splice and you have to deal with the anchorage of the slab's reinforcement. So make sure the splicing doesn't affect interfere with anything else.</p>
<hr>
<p>Now, if you're dealing with a moment-resisting column, then the best splicing location is wherever the moment is minimal, which frequently is near the midspan.</p>
<p>If, for some odd reason it is near one of the slabs, remember to check for interferences.</p>
<hr>
<p>An important exception is if you have to deal with earthquake loads. In this case, one should always try to splice column reinforcement at the midspan. This is because the inertial loads at the nodes are huge, and you don't want to rely on your splice in this delicate area.</p>
<hr>
<p>I also found <a href="https://www.quora.com/Which-is-the-best-zone-to-provide-lapping-in-columns-and-beams" rel="nofollow noreferrer">this discussion</a> on the subject useful.</p>
| 15088 | At What height do we splice vertical bars in Columns and Shear walls? |
2017-05-02T16:57:55.270 | <p>If we have a series of columns as shown in the picture below and to reduce deflection of the slab between two columns that are 11.7 m apart we placed a drop beam 30 x 100 cm (the slab is 35 cm thick) do we need to check the punching shear for these two columns ? and if we need to check the punching shear do we use the beam thickness (100 cm) or we use the slab thickness (35 cm)</p>
<p>I am asking this particular question because in SAFE for columns C5 through C12 does calculate the punching shear</p>
<p><a href="https://i.stack.imgur.com/PMVMt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PMVMt.png" alt="Punching Shear"></a></p>
| |structural-engineering|civil-engineering|design|structural-analysis| | <p>Absolutely.</p>
<p>This can be clearly seen when you consider the load path for everything other than the loads in the region with the beams. Where will a load between C1 and C5 go? Straight to C1 and C5, of course.</p>
<p>A load between C5, C6, C9 and C10 will probably go to the beams and then the columns, so there's no need for punching shear reinforcement in that region, sure, but the loads outside of that area will concentrate right on the columns.</p>
<p>The only question is what the load should be. When calculating the punching shear for column C6, what load should you use?</p>
<p>One option to think about is to just use the load applied in the rectangle bounded by C1-C3-C7-C5 (well, the portion of it that goes to C6). This seems reasonable: this load goes straight to the column (requiring punching shear reinforcement), while the load around C5-C7-C11-C9 first goes to the beams, avoiding the need for punching shear reinforcement. Basically (where full lines represent loads that go straight to the columns and dashed lines go to the beams):</p>
<p><a href="https://i.stack.imgur.com/c5T1v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/c5T1v.png" alt="enter image description here"></a></p>
<p>But how sure can you be that some fraction of the load in C5-C7-C11-C9 won't go straight to the column? After all, the column is far more stiff than the beam, so shouldn't it absorb more of the load than'd be trivially expected?</p>
<p>A really conservative calculation would ignore the beams and calculate the punching shear as if all the load went to the columns. Worst case, you're considering around double the real load.</p>
<p>Another, slightly less conservative calculation would use a 2/3 slope to define the load paths that go straight to the column:</p>
<p><a href="https://i.stack.imgur.com/o2pcw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o2pcw.png" alt="enter image description here"></a></p>
<p>If, however, you actually have a FEM program that can accurately calculate the slab-beam-column interaction, there's an easier way to know the punching force: get the column's axial load and then subtract the shear force of the beam at the column. With this, you'll get the total force that was directly absorbed by the column, which is the punching shear force. Just obviously make sure to model the structure accurately.</p>
| 15091 | Do we need to check punching shear when a drop beam is connecting columns? |
2017-05-03T22:43:38.133 | <p>I'm wondering if the structural integrity of a solid cylindrical pylon is just as sturdy as a cubical one of the same volume given that they're both the same height.</p>
| |structural-engineering|structural-analysis|structures|stresses| | <p>I assume that both pylons are composed of the same material. If you have a cylindrical pylon with height $h$ and volume $V$ then the radius can be calculated by $V = hA_{\text{base}} =h\pi r^2 \implies r = \sqrt{\frac{V}{h\pi}}$. For the cuboid pylon with a square cross-section with a length $a$ for the edge we can calculate $a$ by $V = h A_{\text{base}}=ha^2 \implies a=\sqrt{\frac{V}{h}}$. </p>
<p>Depending on what you mean by sturdy we can distinguish different kinds of loads.</p>
<p><strong>Short summary</strong>: Circular structures are stiffer but more expensive to manufacture and sometimes not suited for the usage (e.g. as floor structure in buildings, where commonly I-beams are used). Read the part below for more detailed answer. </p>
<ol>
<li>Axial load (compressive and tensile in elastic regime) $F$: As $F=\sigma A_{\text{base}}$ both structures have same "sturdyness" because $A_{\text{base}}$ is the same for both cases, which implies $\sigma$ to be the same. But this is only true in theory, because we neglected the effects of the geometry differences and the finite length of the structures.</li>
<li>Torsional load $T$: For this case we have $\tau_{\text{max}}=\frac{T}{W_T}$, in which $W_T$ is the torsional resistance torque. For a circular cross-section, we have $W_T=\frac{\pi}{2}r^3$ and for a square cross section we have $W_T=0.208 a^3$. If you calculate
$$\frac{\tau_{\text{max,r}}}{\tau_{\text{max,a}}}=2\cdot 0.208 \sqrt{\pi}\approx 0.737$$ you will see that the maximal shear stress for the circular structure is preferable, as it is less. Similarly by using $\dfrac{d \varphi}{dx}=\frac{T}{GI_T}$, in which $G$ is the shear modulus (is same for both cases) and $I_T$ is the torsion constant. The integral of the derivative $\dfrac{d \varphi}{dx}$ gives the amount of rotation at length $x$. $I_{T,r}=\frac{\pi}{2}r^4$ and $I_{T,a}=0.141a^4$. Again consider the fraction $$\frac{(d \varphi/dx)_{r}}{(d \varphi/dx)_{a}}=\frac{I_{T,a}}{I_{T,r}}=2\cdot 0.141 \pi\approx 0.886.$$ This implies that the circular cross section will show less rotational deflection unter the same torsional load $T$.</li>
<li>Consider the bending torque $T_y$. The bending line $w(x)$ can be determined for Bernoulli-Euler-beams by solving the following differential equation $$EI_yw''(x)=-T_y.$$ For circular cross-section $I_{y,r}=\frac{\pi}{4}r^4$ and for square cross-section $I_{y,a}=\frac{1}{12}a^4$. If you integrate the differential equation and apply boundary conditions you can calculate the maximal bending / deflection. You could also consider the stress due to $T_y$ which is given by
$$\sigma = \frac{T_y}{I_y}z_{\text{height in beam cross section}}.$$
The fraction $\sigma_{r}$ by $\sigma_{a}$ is given as:
$$\frac{\sigma_r}{\sigma_a}=\frac{I_{y,r}}{I_{y,a}}=\frac{3}{\pi}\approx 0.955.$$ Which indicates that the stress in the circular structure is more favorable.</li>
</ol>
| 15117 | is a cylindrical pylon just as sturdy as a cubical one of the same volume? |
2017-05-04T05:15:01.103 | <p><a href="https://i.stack.imgur.com/VbThd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VbThd.png" alt="enter image description here"></a></p>
<p>Every train car I can remember has its axles arranged like the photo. Two are put very close together in one chassey, and two more the same way at the other end.</p>
<p>But why do that? If you want 4 axles, isn't it better to spread them out evenly like I drew? Or if you just want axles at the ends, why not just have 2 total instead of 4 unevenly spaced ones?</p>
<p><strong>So why are there unevenly spaced axles?</strong> If you want only 2 pivot points, why not just have 2 axles total (which would be spread evenly if they are equal distances from the ends of the car)? If you want 4 axles (presumably for load bearing), why not spread the axles evenly?</p>
<p>I could not find this answer via googling. It might help if I knew the exact terminology of that chassey holding two axles.</p>
| |rail| | <p>The best kind of structure for riding along rails is a pair of axles that are held parallel to each other, and are separated by a distance roughly comparable to the space between the rails. While a single rigid axle containing two conical wheels would self-steer to follow curves if the rails were uniform and free of defects, it wouldn't take much of a defect to knock an axle sufficiently far away from perpendicular to the rails that one or both wheels would cease to be supported by the rails. If there are two axles that are held parallel to each other, and the distance between them is much smaller than any turning radius, such a thing can't happen absent some truly gross defects in the rails.</p>
<p>One could design rail cars with two rigid axles. Such cars were and still are used in Europe. The distance between axles, however, must be small relative to the minimum turning radius the cars will have to traverse, and consequently rail cars that only have two axles will need to be short enough that the portions of the rails near the front wheels are close to parallel with those near the back wheels. Otherwise, if the rails aren't parallel to each other, there would be no way for them to be perpendicular to both axles.</p>
<p>Longer rail cars are constructed with a pair of bogeys that will each ride the rails in the same way as would a very short two-axle rail car. The body of the car is then connected to the center of each bogey. Although the body of the car will of course have considerable mass an inertia, it will not prevent the two bogeys from following the tracks the same way as they would if they were independent vehicles.</p>
<p>The fact that rail cars have four axles is not generally necessitated by the need to distribute the load, but rather by the need to ensure that each axle is held parallel to another axle which is close enough that the portions of the rails near each axle will be nearly parallel. Unless it is necessary to ensure that a very heavy load is spread out e.g. between bridge trestles, increasing the space between the axles on a bogey would make it work less well, without offering any compensating advantage.</p>
| 15121 | Why do train cars have unevenly spaced axles? |
2017-05-04T20:19:24.827 | <p>I am currently working on creating a tool to calculate the maximum heating rate on a re-entry vehicle given starting conditions and coefficients. I am using <a href="https://arc.aiaa.org/doi/book/10.2514/4.862342" rel="nofollow noreferrer">Re-Entry Aerodynamics by Wilbur L. Hankey</a> as my current source. On page 30, he introduces the following equation as a way to calculate the heating rate:</p>
<p><img src="https://chart.googleapis.com/chart?cht=tx&chl=%5Cdot%7Bq%7D%20%3D%20K%20%5Crho%5Em%20V%5E3" alt="Equation" /></p>
<p>The problem is that I do not understand the equation. The way I interpret it is:</p>
<p><img src="https://chart.googleapis.com/chart?cht=tx&chl=%5Ctext%7BRate%20Of%20Change%20Of%20Heat%7D%20%3D%20%5Ctext%7BThermal%20Condcutivity%7D%20*%20%5Ctext%7BDensity%20Of%20Air%7D%5E%7B%5Ctext%7BMass%20OfObject%7D%7D*%5Ctext%7BVelocity%7D%5E3" alt="Equation" /></p>
<p>What doesn't make sense to me is the ρ^m in the equation. If you attempt to solve for the units of ρ^m, you get it as K s^3/m^4, which makes no sense to me.</p>
<p>My researching myself has been unable to lead me to an origin or explanation for the equation, and neither my coworkers or my student friends have ever seen it before. If someone is able to give me some idea what I'm looking at, that would be great.</p>
| |fluid-mechanics|aerospace-engineering|aerodynamics| | <p>$m$ is an empirical curve-fitted constant, depending on the composition of the atmosphere, not a mass. For earth and mars, $m = 0.5$.</p>
<p>Actually the "3" is also a curve-fitted constant. Your reference to Hankey is behind a paywall so I don't know why he fixed that at 3 but left $m$ variable.</p>
<p>See <a href="https://tfaws.nasa.gov/TFAWS12/Proceedings/Aerothermodynamics%20Course.pdf" rel="nofollow noreferrer">https://tfaws.nasa.gov/TFAWS12/Proceedings/Aerothermodynamics%20Course.pdf</a> page 19.</p>
| 15134 | Finding Source of Heating Rate Equation |
2017-05-05T04:51:45.443 | <p>I am doing mechatronics engineering and i am to choose an elective module. We have the choice of taking up Digital Signal Processing or Power Electronics and Drive. Majority of my friends are taking DSP cause our lecturer says its more useful for mechatronic student. I personally feel that PED has a better application for robotics. Could i get some advice on which is more reasonable for robotics.</p>
| |electrical-engineering|power-electronics|signal-processing| | <p>DSP is relevant to any engineering field that involves gathering data and/or controlling something. That includes almost all robotics applications.</p>
<p>It's not quite clear just from the title what "Power electronics and drive" means. "Power electronics" could be anything from a 20W amplifier driving a motor, through a few MW for transport applications (e.g. electric trains), up to GW of power and systems operating at hundreds of kV for electricity generation and distribution.</p>
<p>The bottom line is that working in industry you will almost certainly end up knowing something about both fields, so choosing which to study right now might not be too important in the long run. But I would suggest that DSP would be easier to learn from an academic course, and power electronics easier to learn "on the job" as and when you need it, so long as your degree course has covered the basics of (low power) analog electronic circuit design and analysis.</p>
| 15141 | PED or DSP for mechatronic student |
2017-05-06T18:16:46.177 | <p>I'm a non-engineer doing a little design project.</p>
<p>I'm going to be running some M5-ish (have some flexibility) setscrews into a shaft of about 6mm. There's no reason why the shaft can't be flatted off a bit so I'm presuming I'll do that.</p>
<p>There's no motors or electronics involved and the ends of the shafts are getting a pretty minor (I don't have numbers, sorry, I know I'll need them at some point and will need to do testing to get there) amount of force applied from a distance 20-30mm away. The setscrew connection must be highly resistant to slipping or loosening over extended (could be effectively permanent) use but also must be re-settable occasionally. When it does get reset, it could be right in the area of prior settings on the shaft, and it will have to be done with high accuracy (say down to a resolution of .02mm ideally), so I want to avoid marring as much as possible. (If this isn't possible then the whole thing will have to use a pinch bolt instead of a setscrew.) The shaft can be any material, within reason.</p>
<p>The part will have some exposure to vibration but I don't intuit that it will be more than blue loctite will protect against.</p>
<p><strong>How do I do the math to determine what exact materials to use and the torque on the setscrew to avoid the screw loosening, slipping, or significantly marring the surface of the shaft?</strong></p>
<p>Any recommendations for reading on the topic would be great.</p>
| |torque|bolting| | <p>While 6mm is at the small end of shaft sizes, why not use a standard keyed shaft and hub? The standard key is 2mmx2mm for a 6mm shaft. For a normal fit, the shaft tolerance should be (-0.004/-0029) and the hub (+/-0.012). The setscrew tightens into the key so damaging the shaft becomes a non-issue. A keyed shaft/hub isn't going to slip unless you shear the entire key and if that happens, you've got much bigger problems. The alignment is repeatable to within the tolerance between the width of the key and the keyslots on the shaft and hub.</p>
| 15169 | How to determine materials and torque for a flat setscrew that won't loosen, slip, or mar |
2017-05-07T14:07:03.157 | <p>I am given a step response of magnitude of 3 and the root locus and I have to find the transfer function of the system. The function I find gives me the step response(magnitude of 3 again) of the last diagram.</p>
<p><a href="https://i.stack.imgur.com/oh183.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oh183.jpg" alt="enter image description here"></a></p>
<p>I'm a beginner at this so I've done something stupid probably but I have trouble finding answers regarding control engineering on the internet. This is what I tried doing:
I found the poles and the zeros from the root locus. z=-5,+4 p=-6,-10,-3
I think my transfer function is given from this formula but I'm not sure if we have an H(s) in the feedback and it is not stated : $$ T(s)= \frac{KG(s)}{1+KG(s)} $$
From the poles and the zeros my open-loop transfer function G(s) is :
$$ G(s)= \frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)}$$ Doing the calculations I find : $$ T(s)= \frac{Ks^2+Ks-20K}{s^3+(K+19)s^2+(108+K)+180-20K}$$
From the step response(final value is 4) and the final value theorem I find $\frac{-20K}{180-20K}=-4/3=>K=5.14$ I divided 4 by 3 because the first step response is of magnitude of 3. With this K the step response is the one in the third diagram.It's close to the first one but it's not the one I'm looking for.</p>
<p>What am I missing here? </p>
| |control-engineering|control-theory| | <p>I think you did mix this up with proportional feedback. The transfer function is given by </p>
<p>$$G(s)=K\frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)},$$</p>
<p>in which $K$ is a parameter that needs to be determined.</p>
<p>Because $G(s)$ is a stable plant (all poles are in the left half plane) we can determine the DC gain by the final value theorem.</p>
<p>$$G(s=0)=K\frac{5\cdot (-4)}{10\cdot 6 \cdot 3}=-\frac{K}{9}$$
And $G(s=0)=y(s=0)/u(s=0)=\frac{-4}{3}\implies a=12.$
Hence, </p>
<p>$$G(s)=12\frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)}.$$</p>
<p>MATLAB testing:</p>
<pre><code>s = tf('s');
G = 12*(s+5)*(s-4)/((s+10)*(s+6)*(s+3));
step(3*G); % 3 to scale the unit step response
</code></pre>
| 15189 | Find transfer function from step response and root locus? |
2017-05-08T07:03:52.653 | <p>I am trying to build a "mule": it s basically a go kart chassis with a 12v dc motor (I figured it would be easier to use). Now I need to determine how powerful should the motor be... </p>
<p>It is going to have 4 small wheels (15 cm in diameter - maximum), and be rear wheel driven, and it's going to do some soft kind of off roading (backyard terrain)<br>
It is going to carry up to 150 kg at low speeds (5km/h max). Should I use a torque converter? I was thinking of using 2 cordless drills to do the job... </p>
<p>Do I need torque or power in this case? I'm thinking I'd need plenty of torque as it is a pretty heavy load and the project requires low speeds... </p>
<p>How can I determine the needed power / torque for future projects? Is there a formula (considering air friction 0).</p>
| |electrical-engineering|motors|torque|power| | <p>The force required $F_{\text{r}}$ for driving a wheeled vehicle with mass $m$ is given the following formula (I will neglect slip):</p>
<p>$$F_{\text{r}}=F_{\text{tire}}+F_{\text{aero}}+F_{\text{acc.}}+F_{\text{slope}}.$$</p>
<p>The frictional force of the tires is given by $F_{\text{tire}}=c_{\text{tire}}mg.$ The dimensionless tire friction coefficient $c_{\text{tire}}$ is in general between $0.005$ and $0.010$.</p>
<p>Aerodynamic resistance is given by $F_{\text{aero}}=\frac{1}{2}\rho_{\text{air}}c_{\text{D}}A_{\text{ref}}v^2_{\text{rel}}.$ The dimensionless drag coefficient is between $0.28$ (good aerodynamics) and $0.80$ (bad aerodynamics e.g. for trucks). The reference area $A_{\text{ref}}$ is the projected area of the front face of the vehicle. For very small relative velocities $v_{\text{rel}}=v+v_{\text{wind}}$ aerodynamical resistance can be neglected.</p>
<p>The force needed for a given translative acceleration $\ddot{x}$ is given by $F_{\text{acc.}}=e_{\text{m}}m\ddot{x}$. The dimensionless coefficient $e_{\text{m}}$ is there to account that it is also necessary to accelerate the components of the motor, gearbox and so forth. In most cases it is between $1.05$ and $1.40$.</p>
<p>The force $F_{\text{slope}}=mg\sin{\alpha}$ is necessary to overcome a slope of $\alpha$ (in degrees, so remember to calculate $\sin{\alpha}$ in degrees and not in rad). </p>
<p>So in total we get:</p>
<p>$$F_{\text{r}}=c_{\text{tire}}mg+\frac{1}{2}\rho_{\text{air}}c_{\text{D}}A_{\text{ref}}v^2_{\text{rel}}+e_{\text{m}}m\ddot{x}+mg\sin{\alpha}.$$</p>
<p>in order to get the required power $P_{\text{r}}$ you simply multiply $F_{\text{r}}$ with the velocity $v$ of the vehicle. Note that in general, when considering the wind this velocity $v$ will not be equal to the relative velocity $v_{\text{rel}}$. </p>
| 15208 | How to calculate the power/torque/force needed to move a 4 wheeled vehicle? |
2017-05-08T08:34:46.840 | <p>I have a beam as shown below that have a span having 60 cm in thickness and the other spans 30 cm do I consider this beam as continuous or does the changing in thickness affect that ? if there is anything like an article about these criteria (especially in ACI code)
<a href="https://i.stack.imgur.com/EyuPj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EyuPj.png" alt="Span 3 with changing thickness"></a></p>
| |structural-engineering|civil-engineering|design|structural-analysis| | <p>Yes it is still continuous; all beams are still capable of producing a hogging moment above the columns.</p>
| 15209 | What is the thickness criteria to consider a beam with changing thickness Continuous? |
2017-05-09T06:29:55.813 | <p>In the below picture I want to design B4 and B6 but I am not sure which one does support the other and how the detailing of the area of intersection between these two beams?</p>
<p>Is it better that B6 supports B4 because B6 have the shortest span?</p>
<p><a href="https://i.stack.imgur.com/XULmT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XULmT.png" alt="Area intersection between two beams"></a></p>
| |structural-engineering|civil-engineering|design|structural-analysis| | <p>You decide which beam supports which by how you detail the steel reinforcement at the intersection.</p>
<p>Usually, one chooses to make the stiffest beam support the other. In your case, that means that yes, B6 would support B4. However, that decision isn't taken merely by looking at the beams' spans, but also at their $EI$. Since they seem to be of same (or very similar) cross-section, then the difference in span becomes the only relevant variable. However, if B6 were actually half as wide (or tall, which would be even worse), then it would probably be far less stiff than B4, in which case one'd define that B4 supports B6.</p>
<p>If you prefer, you can also let an analytical model make that decision for you. Look at the shear diagrams of both beams at the intersection. It should be clear that one "lost" shear force and the other "gained" it, meaning the latter is supporting the former.</p>
<hr>
<p>Now, once you've defined which beam supports which (in this case, B6 supports B4), we need to inform our decision to the universe by detailing that intersection accordingly.</p>
<p>It is not merely a matter of which beam's rebar goes beneath the other's. After all, it is common to assume that reinforced concrete transmits its load from the bottom of the beam, at the end of the compression strut:</p>
<p><a href="https://i.stack.imgur.com/ylo4t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ylo4t.png" alt="enter image description here"></a></p>
<p>This means that, as far as the supporting beam is concerned, it just received a concentrated load near its bottom fiber. Therefore, additional suspension reinforcement in the form of stirrups is necessary to "raise" this load, allowing the full height of the supporting beam to work. This reinforcement is found by obtaining the support reaction between the beams and calculating</p>
<p>$$A_s = \frac{R}{f_{yd}}$$</p>
<p>(when the supported beam's bottom fiber is above the supporting beam's bottom fiber, this reinforcement can be reduced somewhat). This reinforcement should be added to the normal stirrups and doesn't need to be exclusively within the intersection volume. Item 9.2.5 of Eurocode 2 allows the suspension stirrups to be placed in the following region:</p>
<p><a href="https://i.stack.imgur.com/zdtO7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zdtO7.png" alt="enter image description here"></a></p>
| 15232 | How is the detailing of the intersection between two RC beams perpendicular to each other? |
2017-05-09T15:43:53.630 | <p>I'm working on a project trying to combine a <a href="http://www.mekanizmalar.com/pantograph.html" rel="nofollow noreferrer">pantograph linkage</a> with a <a href="http://www.mekanizmalar.com/four-bar-infinity-curve.html" rel="nofollow noreferrer">four-bar linkage</a>. I'd like to build it out of relatively at-hand materials (AKA, nothing requiring a machine shop), and I'd like to do some educational projects in the future involving students building linkages, too, so low-cost would be great. Relatively stiff and able to run unattended as a display are also desirable. </p>
<p>I'm thinking that bolts and nuts are going to be problematic after it runs for too long. Is there a particular type of bearing that would be easy to use/acquire? ...a good material for the rods? I have a Carvey CNC machine, so HDPE and wood are easy to machine for me. It's mostly the pivots/bearings that I'm having trouble with.</p>
| |bearings|linkage| | <p>I have tested linkage models using wood strips with small bearing recesses milled into the wood. Skate bearings 8mm ID x 22mm OD x 7mm width are widely available for about 60 cents if you buy 20 or so. I just milled the bearing recess a bit tight then pressed them in. If those are too large then smaller bearings are also available and may be cheaper in bulk. Be aware that skate bearings are available in some exotic ceramics and other materials that get very expensive, so don't be scared off if you see some for $10 per bearing. I have paid about 40 cents for them when I bought them in bulk.</p>
| 15239 | What are a few low-cost options for building linkages? |
2017-05-09T20:15:15.403 | <p>If I have a rectangular embedded beam in a slab supporting hollow core slab how is the connection and detailing between these two elements ? (any sketch would help)</p>
<p>If the beam parallel to the hollow core how is the connection between these two ?
<a href="https://i.stack.imgur.com/WHGqL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WHGqL.png" alt="enter image description here"></a></p>
<p>I need to know how the connection between the hollow core and the beam how is the bars between these two are connected ?</p>
| |structural-engineering|civil-engineering|design|structural-analysis| | <p>This depends on what connection behavior you would like to achieve. The hollow core slab manufacturer will typically have standard connection details or assist you with the connection design. Usually beams/columns are designed such that the prefabricated hollow core slabs can be lifted into place. Therefore, usually there are corbels or ledges for the slabs to sit on. </p>
<p>To get an idea, Hollow Core Concrete Pty Ltd (Australia) has <a href="http://www.hollowcore.com.au/floor_slab_tech.php" rel="nofollow noreferrer">Hollow Core Slab Technical Sheets</a> on their website showing some typical connection details.</p>
<p>For example, I have taken the detail for an <a href="http://www.hollowcore.com.au/pdf_files/Internal%20Beam%20to%20Hollow%20Core%20Connection.pdf" rel="nofollow noreferrer">internal beam to hollow core connection</a> from their website: <a href="https://i.stack.imgur.com/Us41j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Us41j.png" alt="internal beam to hollow core connection"></a></p>
<p>It is worth discussing the design with the hollow core manufacturer to get an expert opinion. I always recommend using the experts instead of trying to re-invent something.</p>
<p>As @Wasabi pointed out I didn't address the situation you mention in the comments with the slab and beam having the same depth. This is a non-standard detail with regards to hollow core slabs. As I mentioned earlier, hollow core slabs are typically used in pre-cast construction. If you are casting integral beams, this removes much of the advantage of using hollow core slabs in the first place. You could just cast a <a href="http://www.concreteconstruction.net/how-to/voided-slab-flat-plate-floor-construction_o" rel="nofollow noreferrer">voided slab</a> in-situ for example. </p>
<p>If you need to use hollow core slabs in this situation there are potentially a few ways this could be achieved, for example: breaking out part of the hollow-core slab, adding the additional reinforcement then casting new concrete to fill the space; or using a <a href="http://www-civ.eng.cam.ac.uk/struct/jml/CCSG_topic_prhd2.html" rel="nofollow noreferrer">concrete half-joint</a> detail. The manufacturer may also be able to incorporate this beam into the design of the slab. Any of these non-standard connections would depend on your loading and other design and construction constraints and would likely need to be designed in collaboration with the hollow core slab manufacturer.</p>
<p>If you are willing to use steel-concrete composite construction you could use a typical slimfloor approach:</p>
<p><a href="https://i.stack.imgur.com/OsT6jm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OsT6jm.jpg" alt="slimfloor"></a></p>
<p>The space around the steel beam would be filled with concrete, and if you wanted to get very clever you could even add rebar in this space.</p>
| 15243 | How the connection is between a Hollow core slab and an embedded beam? |
2017-05-11T12:53:28.977 | <p>I am trying to translate the name of a special kind of beam from a set of Russian-language drawings. </p>
<p>The dictionary translates the name as simply supported beam, but it's not that simple. There is a slight difference in two kinds of beams. First - what is called a <strong>simply supported beam</strong>:</p>
<blockquote>
<p>A beam that lies freely on two supports and covers one span. </p>
</blockquote>
<p>Second - <strong>my beam</strong>. This one is different from the previous that it lies on multiple supports and covers many spans.</p>
<p><em>Important note: both of the beams are not continuous; i.e., they are parts of one long beam.</em></p>
<p>The question is, what would I call the second beam?</p>
<p><a href="https://i.stack.imgur.com/0WkKu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0WkKu.jpg" alt=""></a></p>
| |structures|beam|terminology|building-design| | <p>This is also considered a simply-supported beam. If you want to be more descriptive, you can call it a simply-supported beam with overhangs (the cantilevers beyond the main span).</p>
<p>Basically, any beam which rests only on two hinged supports (allowing rotations) is considered simply supported. Beams which rest on more than two supports are considered continuous (so long as all the spans are covered by the same beam):</p>
<p><a href="https://i.stack.imgur.com/QM7oY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QM7oY.png" alt="enter image description here"></a></p>
| 15262 | Translating the name of a type of beam from Russian to English |
2017-05-12T00:02:54.910 | <p>I am trying to transform any arbitrary vector $\bf{v_{1}}$ in a Cartesian frame $\mathscr{F_{1}}$ into a different Cartesian frame $\mathscr{F_{2}}$. Both $\mathscr{F_{1}}$ and $\mathscr{F_{2}}$ have the same origin. I have consulted <em>Aircraft Flight Dynamics and Control</em> by Wayne Durham, and understand that book's solution as follows:</p>
<p>Use the standard yaw-pitch-roll $ z,y,x$ $/$ {$\theta_{z},\phantom{s} \theta_{y},\phantom{s} \theta_{x}$} (usually called {$\psi, \theta, \phi$} respectively) / $3 2 1$ order. There exist three transformation matrices, one for each axis. First, rotate about the z axis of $\mathscr{F_{1}}$. Call the transformation matrix $T_{\mathscr{F'},1}$ since it goes from $\mathscr{F_{1}}$ to an intermediate frame $\mathscr{F'}$. $T_{\mathscr{F'},1}$ is given as </p>
<p>$$T_{\mathscr{F'},1}=\begin{bmatrix}
cos(\theta_{z}) & sin(\theta_{z}) & 0 \\
-sin(\theta_{z}) & cos(\theta_{z}) & 0 \\
0 & 0 & 1 \\
\end{bmatrix}$$</p>
<p>To go from $\mathscr{F'}$ to $\mathscr{F''}$, you then rotate about the y' axis using
$$T_{\mathscr{F''},\mathscr{F'}}=\begin{bmatrix}
cos(\theta_{y}) & 0 & -sin(\theta_{y}) \\
0 & 1 & 0 \\
sin(\theta_{y}) & 0 & cos(\theta_{y}) \\
\end{bmatrix}$$</p>
<p>Finally, to go from $\mathscr{F''}$ to $\mathscr{F_{2}}$, you then rotate about the x'' axis using</p>
<p>$$T_{\mathscr{F''},\mathscr{F'}}=\begin{bmatrix}
1 & 0 & 0 \\
0 & cos(\theta_{x}) & sin(\theta_{x}) \\
0 & -sin(\theta_{x}) & cos(\theta_{x}) \\
\end{bmatrix}$$</p>
<p>A composite transformation matrix $\bf{T}$ can be made by multiplying these transforms together. Individual transform steps on a generic vector $\bf{v}$ are given in the book (and I reproduced them below) to show order determination. </p>
<p>$$\bf{v'}=T_{\mathscr{F'},1} \bf{v}$$
$$\bf{v''}=T_{\mathscr{F''},\mathscr{F'}} \bf{v'}$$
$$\bf{v_{2}}=T_{\mathscr{F_{2}},\mathscr{F''}} \bf{v''}$$</p>
<p>$$\bf{v_{2}}=T_{\mathscr{F_{2}},\mathscr{F''}} T_{\mathscr{F''},\mathscr{F'}} T_{\mathscr{F'},\mathscr{F_{1}}} \bf{v}$$</p>
<p>$$\bf{v_{2}}=\bf{T}v_{1}$$</p>
<p>So $\bf{T} = T_{\mathscr{F_{2}},\mathscr{F''}} T_{\mathscr{F''},\mathscr{F'}} T_{\mathscr{F'},\mathscr{F_{1}}}$ (strictly in this order).</p>
<p>My problem arose when I coded this transformation into MATLAB and tested some examples which I could verify qualitatively (with some coordinate axes I made out of wood).</p>
<p>The transformation matrix $\bf{T}$ that works is $\bf{T} = (T_{\mathscr{F'},\mathscr{F_{1}}} T_{\mathscr{F''},\mathscr{F'}} T_{\mathscr{F_{2}},\mathscr{F''}})^{T}$. Thus, in order for the transformation to work the way I expect it should, I need to multiply the matrices in reverse order and take their transpose. What has gone wrong?</p>
| |control-engineering|aerospace-engineering|mathematics| | <p>You have actually not done anything wrong. Its just that Matrix algebra, much to the chagrin of some, does not actually define how you need to pack your data into the matrix. Your choice.</p>
<p>So the question is do you consider that vectors inside your matrix to be columns or rows. This is called column major or row major. What you simply get is a transposed solution, and a inverse calculation order. It happens, some literature is in row major and some is in column major notation. Its all a question of whether you you want to look form object out or form outside in, your choice. But you need to be aware of this or you end up with problems like this, or well they aren't problems just definitinons.</p>
<p>Using qaternions or any other method does get you the same issue.</p>
| 15272 | 3D Transformation Between Two Cartesian Coordinate Systems Using Euler Angles |
2017-05-12T14:07:18.353 | <p>All,</p>
<p>I'm designing a minor slab to Eurocode 2. It's purpose is more for show than any real structural purpose however I need to justify it's ability to resist it's own self weight and possibly the accidental loading of a person climbing upon it.</p>
<p>The slab itself is 680mm wide, 160mm deep (100mm effective depth to rebar) and has a maximum span of 1200mm, so there is little room for shear reinforcement. Ideally, this slab should be kept as small and simple as possible. </p>
<p>This is a small slab and acts as a cope for a sheepile wall and so spans between the 'zig-zag' of the piles. </p>
<p>Eurocode 2 states the following: </p>
<p><a href="https://i.stack.imgur.com/1IcM7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1IcM7.png" alt="Clause 1, Section 6.2.2, Eurocode 2 (BS EN 1992-1-1)"></a></p>
<p>As can be seen, for slabs with an effective depth (<em>d</em>) of less than 200mm, the equations do not hold up as the inequality K <= 2.0 must be satisfied.</p>
<p>There is very little explanatory text around these clauses and I am looking for guidance. Does this mean that a solution is impossible using these equations? If so, what other guidance can I refer too?</p>
<p>If not, what significance does the k value have? Does a k value > 2.0 give me a conservative result that I can continue to use?</p>
<p>Ultimately; How can the shear capacity of a concrete slab be determined for effective depths under 200mm within Eurocode 2? </p>
| |structural-engineering|civil-engineering|concrete|standards|eurocodes| | <p>As @Wasabi and @AndyT mentioned in the comments, this means the maximum value $k$ can have is $2$, so if $d<200$ then $k=2$. This can be confirmed in the section on shear design for slabs from the document <a href="http://www.concretecentre.com/Publications-Software/Publications/How-to-Design-Concrete-Structures-to-Eurocode-2.aspx" rel="nofollow noreferrer">'How to Design Concrete Structures to Eurocode 2'</a>. Table 7 from this document shows:</p>
<p><a href="https://i.stack.imgur.com/yXJln.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yXJln.png" alt="enter image description here"></a></p>
<p>which has $V_{Rd,c}$ unchanged for effective depths $d<200$ and $k=2$ at the bottom of the $d\leq 200$ column. You can confirm this by doing the calculation by hand. Using $k>2$ would be non-conservative as it would predict a higher shear capacity than allowed by the code.</p>
<p>If you are interested in a more complete discussion of where this formula comes from you could refer to the <a href="http://www.europeanconcrete.eu/publications/eurocodes/114-commentarytoeurocode2" rel="nofollow noreferrer">Commentary to Eurocode 2</a>, essentially it is a fit to experimental data.</p>
| 15278 | How can the shear capacity of a concrete slab be determined for effective depths under 200mm? [Eurocodes] |
2017-05-12T14:50:00.817 | <p>I've run across a term in some grinding manufacturing research papers called the <em>specific material removal rate</em>. The units given for this term are [mm$^3$/mm$\cdot$s] (I don't really like this style of formatting because it makes it unclear as to whether the [s] is in the numerator or denominator, but that is the style that has been used in all of the papers that I have seen with this term). I am familiar with the general machining term <em>material removal rate</em> (or $MRR$), which is the volume of material removed per second from a workpiece in [mm$^3$/s], but what is the "specific" part of the "specific material removal rate"?</p>
<p>Usually in machining when we say "specific" we're talking about per unit volume removed. So the <em>specific energy of machining</em> is normally expressed as unit energy per unit volume removed (e.g. [J/mm$^3$]). But for the <em>specific</em> $MRR$ it is a volume per second per unit length (after all, it wouldn't make any sense to have volume removed per unit volume removed). So I guess the real question is, what is this length that we are dividing the $MRR$ by to get the <em>specific</em> $MRR$? Why is this term useful?</p>
<p>This seems like a pretty basic question, but I have been unable to find a definition for the specific material removal rate anywhere. I've looked at several papers and from context in the papers it is rather unclear. I've checked my textbook, <em>Fundamentals of Modern Manufacturing 5th Ed.</em> by Mikell P. Groover, no answer there. It may be a term that is specific to grinding operations.</p>
<p>I would appreciate it if any answers could be backed up with a source.</p>
| |manufacturing-engineering|grinding| | <p>"Specific Metal Removal Rate, or SMRR, represents the rate of material removal per unit of wheel contact width and are commonly recommended from:
200 to 500 mm<sup>3</sup>/mm width/minute (0.3 to 0.75 in<sup>3</sup>/inch width/minute)."</p>
<p>I know this is an old question but I figured, since there's some confusion, I'd share a direct quote from "Machinery's Handbook" 30th edition, by Erik Oberg, Franklin D. Jones, Holbrook L. Horton, and Henry H. Ryffel. 2016 Industrial Press.</p>
| 15279 | Definition of the "specific material removal rate"? |
2017-05-13T05:47:35.947 | <p>I'm working on a process flow currently and am using carbon dioxide as the working fluid because of its heat properties and because it's an inert gas. In the reactor that I am designing I'm looking to operate at 4 atmospheres (~58.78 psia) and at around 977°F (~500°C). I know it would be a supercritical fluid at this P-T relation but can not find any information on the properties of carbon dioxide at these conditions.</p>
<p>How would I go about finding or calculating for the thermodynamic properties at these conditions for carbon dioxide and then be able to continue this calculation through the entire process I'm designing?</p>
| |fluid-mechanics|thermodynamics|pressure|temperature| | <p>Coolprop is a free API that's very similar to RefProp (not free):
<a href="http://www.coolprop.org/contents.html" rel="nofollow noreferrer">http://www.coolprop.org/contents.html</a></p>
| 15288 | How to find properties of supercritical CO2 as a working fluid in a system |
2017-05-14T20:58:09.760 | <p>If we are able to create a fusion reactor so there is a net gain in energy how can we turn the heat into electricity.</p>
<p>I know how it is done normally for fission, steam, but how could this be adapted for a fusion reactor where all of the heat is confined?</p>
<p>Are there any concepts for the stage after fusion becomes net positive?</p>
| |electrical-engineering|power-engineering|generator| | <p>Thermal energy is, in part, obtained or collected in the breeder blanket concept. Here <a href="https://www.researchgate.net/publication/263201067_A_Thermal_Discrete_Element_Analysis_of_EU_Solid_Breeder_Blanket_subjected_to_Neutron_Irradiation" rel="nofollow noreferrer">neutrons with high energy are used to produce heat and breed valuable tritium</a> by interaction with lithium ceramic breeder pebbles. From interaction with a high energy neutron, Lithium fissions into helium and tritium. The lithium ceramics (usually <a href="https://www.researchgate.net/publication/266857952_Solution_based_synthesis_of_mixed-phase_materials_in_the_Li2TiO3-Li4SiO4_system" rel="nofollow noreferrer">Li4SiO4 or Li2TiO3 or mixtures thereof</a>) are present in the form of a pebble bed, cooled by helium. Heat, tritium and helium are produced within the bed, thus providing further fuel and useful heat for the generation of electricity.</p>
<p>Basically to sum it up, the neutrons emitted in the fusion reaction serve to deliver energy to the external system that is isolated from the plasma (in which the fusion takes place).</p>
| 15307 | Extracting power from fusion? |
2017-05-15T11:25:38.043 | <p>I'm currently using 400ml <a href="http://www.farnell.com/datasheets/316910.pdf" rel="nofollow noreferrer">Dust Off 67</a> gas dusters and want to replace them with compressed air and a blow-out gun.</p>
<p>In order to select the right equipment and get a better understanding I would like to compare the amount of air the gas duster contains to the compressed air solution. I know the gas duster does not actually contain compressed air, but for me it's all about the result. I know how much I can work with the gas duster and want to have a comparison.</p>
<p>So considerung the 400ml Dust Off 67 (see linked datasheet for additional details, I don't know what might be relevant there). If I had a gas container which I can fill up with a pressure of up to 8 bar, how much volume (in liters) would the gas container need to get the same amount of air out of it as I would get the with the gas duster?</p>
<p><a href="https://i.stack.imgur.com/zi0B9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zi0B9.png" alt="enter image description here"></a></p>
| |compressed-air|compressed-gases| | <p>According to to the SDS found <a href="http://www.tme.eu/en/Document/bbd0a19d0233a00349282ad342bbaaed/DUST_OFF_67_Aerosol.pdf" rel="nofollow noreferrer"><strong>here</strong></a> the product contains tetrafluoroethane as an ingredient. I looked around and found other similar products with the same ingredient.</p>
<p>For sake of simplicity lets say that all the $400 ml$ are tetrafluoroethane and the compressed density is $1.01 g/cm^3$ according to your table. With this information we know that one can contains $404 g$ of tetrafluoroethane by multiplying the density by the volume. ( $1 ml = 1 cm^3$ ) According to <a href="https://en.wikipedia.org/wiki/1,1,1,2-Tetrafluoroethane" rel="nofollow noreferrer"><strong>Wikipedia</strong></a> the molar mass of tetrafluoroethane is $102.03 g/mol$. With this information we know that bottle contains 3.96 moles of tetrafluoroethane. ( $404 g / 102.03 g/mol = 3.96 mol$ )</p>
<p>Now we can use the <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/idegas.html" rel="nofollow noreferrer"><strong>Ideal Gas Law</strong></a> to solve for volume at <a href="https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure" rel="nofollow noreferrer"><strong>STP</strong></a>. In our case a handy rule of thumb is 1 mole = $22.4 L$ at STP. So you canister contains 88.7 liters of tetrafluoroethane ( $3.96 * 22.4$ ) if it were not compressed.</p>
<p>Now we reverse this process for air. This time around the handy rule is PV( initial ) = PV ( final ). 8 bar = 7.9 atmmospheres. $1 atm * 88.7 L = 7.9 atm * 11.2L$. So the answer to your question is that you will need 11.2 liters of compressed air at 8 bar to equal to your .4 liters of Dust Off! The reason that Dust Off is so efficient is that it liquefies when compressed where air does not.</p>
<p>If the process seems confusing read up on the Ideal Gas Law. It is very powerful when dealing with gasses at different conditions.</p>
| 15322 | volume comparison of compressed air and gas duster |
2017-05-15T18:04:45.413 | <p><a href="https://i.stack.imgur.com/W6Xz4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W6Xz4.jpg" alt="enter image description here" /></a></p>
<p>Default Hydraulic solenoid valve issue</p>
<p>I'm having a weird issue with some solenoid valves in a simple hydraulic circuit, let me try and describe the situation as best as I could.</p>
<p>Coming out of a 2 station manifold, I have 2 4-way spring return center drain valves. 120VAC, controlled through relay outputs by a PLC. Each valve controls a set of cylinders, A position down, B position up. Wired up such that 1 leg of each coil jumped together to my 120VAC control, the other 4 legs to their corresponding PLC output.</p>
<p>Initially, I had valves with a center position, A&B drain to T, P blocked. Everything worked correctly from a controls standpoint, only problem was when both valves were "off", I was holding at full pressure (~1700 psi)and building up a lot of heat.</p>
<p>So I swapped out the valves for center position, A to B blocked, P drain to T. Wired up exactly the same as before. Now when I actuate, let's say, valve 1, side A, nothing happens...until I actuate valve 2 side A, then cylinder on valve 1 actuates, as well as the cylinder on valve 2. Electrically everything is right when I check with a meter, program is running as it should as well. So that leads me to believe that something's going on hydraulically.</p>
<p>Anyone have any idea what's going on? It seems like when I have one valve, P going to A and the other "off" (P to T) that I might have to much pressure going to tank, preventing B to T thus keeping my cylinder from moving?</p>
<p>Let's assume that's the issue, how can I modify my design to remedy the problem? I'm thinking, change from one, 2 station manifold to 2, independent station blocks and add check valves to the T outputs to prevent pressure from backing into the T line and preventing the oil from draining from the exhaust of the cylinder.</p>
<p>Solved: A "dump" or relief valve was added to the circuit, right before the solenoid manifold, plumbed into the tank line, right before the radiator. In the instances when the pump is running, but the machine/solenoid valves are not operating, the dump valve is opened, draining all pressure to tank through radiator.</p>
| |mechanical-engineering|hydraulics| | <p>Yes. You are losing your pump pressure through the center (off) position of the the other valve. Once you do some work through the other valve, so as not to drain your pressure to the tank, then both cylinders get some pressure to operate. </p>
<p>Here are two approaches. </p>
<p>1) Connect the T output from one valve to the P input of the other valve. The fluid can return to the tank easily when both valves are in the center position, but both valves will receive pressure from the pump. The pressure and volume going to the second valve will not be constant when the valve that is first in line is being switched, but both will still work.</p>
<p>2) I have seen this approach, but I don't know the exact circuit. Go back to your earlier approach, and add a pressure sensing pilot controlled relief valve that opens the return path to the tank whenever neither of your control valves is drawing fluid. So your full 1700 psi drops to some much lower pressure to reduce heat when the pressure is not being drawn by either valve.</p>
| 15334 | Hydraulic System design help/ directional valves |
2017-05-16T13:55:45.070 | <p>I'd like to ask you if a C mex s-function model in Simulink can get the dSpace execution time lower, rather than having a model with blocks and functions.</p>
| |electrical-engineering|control-engineering|matlab|simulink| | <p><strong>Short Answer:</strong> It depends on what blocks you're replacing, and even then the effects might be negligible.</p>
<p>An interconnected set of standard Simulink blocks (gain, sum, integral, derivative, etc.) will definitely be equally-fast or faster than any custom function you write, CMEX or not. On the other hand, replacing a MATLAB Function Block with a CMEX S-Function might have some effect, but it's very difficult to determine.</p>
<p>The effects of using a CMEX S-Function will depend on your skill at writing efficient C code, and even then the effect could be small compared to the execution time of other parts of your model.</p>
<p>It's also not as simple as asking which block is more efficient in 'Normal mode' Simulink. dSpace uses a combination of the Simulink Coder and their own proprietary software to translate your Simulink model into C code that is then compiled into an executable. That executable is then run in real time by the dSpace hardware. Therefore, during the translation process from Simulink to C code the benefits of using the CMEX S-Function might be negated.</p>
<p>*Note: I have not worked with dSpace but I do use a very similar software package in my research called QUARC, which is basically the Canadian equivalent.</p>
| 15345 | dSpace overrun issue |
2017-05-16T22:50:05.560 | <p>We have large scale aircraft with long endurance and much higher speed than birds. But it seems that aircraft of comparable size to birds (i.e. drones) have much lower endurance, top speed and flight range. What technologies are currently limiting us from achieving this?</p>
| |mechanical-engineering|aircraft-design| | <p>Somebody needs to design a chocolate-powered drone. Yes, seriously.</p>
<p>The total energy stored in a 40AH 12V battery is about the same as the calorific value of five 100g chocolate bars from your nearest supermarket. </p>
<p>Source: from <a href="https://www.tesco.com/groceries/product/details/?id=254381873" rel="noreferrer">https://www.tesco.com/groceries/product/details/?id=254381873</a>, "Tesco Everyday Value Milk Chocolate" provides 3840 kJ/kg, and the fully charged battery holds 40x12x3600/1000 = 1728 kJ</p>
<p>Find a way to pack that much energy density into drone (and also consider that unlike LIPO batteries, chocolate bars don't spontaneously combust if they are mishandled!) and the comparison between drones and birds would come out rather differently.</p>
<p>Of course, soaring birds (and sailplanes) can stooge around all day and night, getting their energy from thermals and the wave airflows over mountain ridges - but they still need <em>some</em> external power to reposition themselves to take advantage of those "free" energy sources.</p>
| 15353 | What technologies prevent drones from being as efficient as birds? |
2017-05-17T18:45:27.497 | <p>I was reading <a href="http://www.control.tu-berlin.de/images/2/22/seel2012_msc.pdf" rel="nofollow noreferrer">this article</a> called "<em>Joint Axis and Position Estimation from Inertial Measurement Data by Exploiting Kinematic Constraints</em>", but I'm not understanding one part, which doesn't let me proceed in the reading of the same.</p>
<p>From the physical point of view, we have the following situation:</p>
<p><a href="https://i.stack.imgur.com/IxNHu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IxNHu.png" alt="enter image description here"></a></p>
<p>Two segments (one orange and the other green) connected by a <em>hinge joint</em>. The 3-axis coordinate systems that you see in the picture (the one on segment orange and the other on the segment green) are the local coordinate systems of two 3-dimensional gyroscopes.</p>
<p>The dashed line that you see in the hinge should be the coordinate axis of the same joint. Since the joint can only move in one dimension, we have just one axis.</p>
<p>Now, given this setting, the authors say (first page, column at the right, at the end of the same):</p>
<blockquote>
<p>... let the angular velocities of the gyroscopes, in the coordinates of
their local frames, be $g_1(t)$ and $g_2(t)$ for the first and the second segment, respectively. Then it is a geometrical fact, that $g_1(t)$ and $g_2(t)$ differ only by the joint angle velocity and a (time-variant) rotation matrix</p>
</blockquote>
<hr>
<h3>Questions</h3>
<ol>
<li><p>What is the joint angle velocity?</p></li>
<li><p>What is this time-variant rotation matrix?</p></li>
<li><p>Why do the angular velocities differ only by the joint angle velocity and a time-variant rotation matrix? What does it even mean, actually?</p></li>
</ol>
| |mathematics|terminology| | <p><a href="https://i.stack.imgur.com/CLLFS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CLLFS.jpg" alt="enter image description here"></a></p>
<blockquote>
<ol>
<li>What is the joint angle velocity?</li>
</ol>
</blockquote>
<p>The joint angle velocity is the angular velocity of you joint, because we are acutally dealing with a 2D problem. This will become clear if you look at your system from the $z_0,z_1$ axis direction. In the picture this angular velocity is given by $\dot{\beta}$.</p>
<blockquote>
<ol start="2">
<li>What is this time-variant rotation matrix?</li>
</ol>
</blockquote>
<p>You have to consider four coordinate frames. The first coordinate frame is fixed at your joint axis $K_0=(x_0,y_0,z_0)$. The second frame is frame 1 but now rotating around the $z_0=z_1$ axis. The other two frames are there in your inital image. If you tranform from frame 1 to the other frames you will see that the rotation matrix will be time dependent. This is obvious, because depending on the time dependent motion of $\beta$ the coordinate frames will also move with time dependence.</p>
<blockquote>
<ol start="3">
<li>Why do the angular velocities differ only by the joint angle velocity and a time-variant rotation matrix? What does it even mean,
actually?</li>
</ol>
</blockquote>
<p>This means that the angular velocity of the left coordinate frame $\omega_1$ and the angular velocity of the right coordinate frame $\omega_2$ are related by the angular velocity of the joint $\dot{\beta}$. The relationship is given by:</p>
<p>$$\omega_2-\omega_1=\dot{\beta}.$$</p>
<p>Note, that this simple equation is only true because we only have a 2D rotation. Imagine $\dot{\beta}=0$, then the joint is acting like a fixed link, hence the left and right coordinate frame have the same angular velocity. If $\dot{\beta}$ is positive then the angular velocity of the right frame $\omega_2$ will be the sum of the angular velocity of the left frame $\omega_1$ and the angular velocity of the joint $\dot{\beta}$.</p>
| 15360 | Two angular velocities on two segments connected by a hinge joint differ only by the joint angle velocity |
2017-05-18T04:12:55.570 | <p>I need design a trolley with four casters, the four casters are installed on the four corners of a 460mmX405mm metal plane. And the load will be a cubic 380mmX225mmX350mm (Height is 350mm), weight about 150Kg, site at the center of the metal plane.</p>
<p>The dimensions of the casters is as below:</p>
<p><a href="https://i.stack.imgur.com/hYlKp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hYlKp.png" alt="enter image description here"></a></p>
<p>You see the support point of the caster and the wheel are not on a vertical line, so I'm afraid when all four wheels in the same direction, the trolley will not be stable. So I go here for some suggestions(or some formulas are better), if the trolley can works stable? Or I must change my design, such as enlarge the metal plane? </p>
| |applied-mechanics| | <p>As long as the worst case (all wheels pointing to the center) is still stable then you have no issue.</p>
<p>To look at the footprint first create it as if you didn't use castors. Then for cornet with a castor move the line so it's tangential to the inside of the circle the contact point makes when swiveling. Note that this may mean that the apparent contact point is outside that circle. For a rectangular base you can subtract twice the radius of the swivel's offset on each side.</p>
<p>Then you can use that footprint for your stability calculations.</p>
| 15367 | Design a trolley with four casters |
2017-05-18T07:40:37.413 | <p>I have a small pump that does 6L/min flow-rate. With a pipe size of 0.5 inches and 3 nozzles spaced at a distance of about 1m each, is it possible to do effective misting? My total pipe length will be 4m. I have been trying out calculations from various references but it keeps driving me nuts. What am wondering is, is the pump fit for that job? I have seen similar pumps online with equal flow-rates and pressure ratings of about 130psi. What matters most?</p>
| |mechanical-engineering|fluid-mechanics| | <p>Droplet size through a spray nozzle is a strong function of pressure drop. The higher the pressure drop, the smaller the drop size. So for an atomizer you need high pressure (probably >15 psi). Coarser droplets can be produced down to about 3 psi.</p>
<p>Check out the data sheets for you nozzles. They likely specify acceptable pressure drops and may also rate the droplet size (coarse to fine).</p>
<p>And, as is usual, the pressure drop across the nozzle is related to flow rate. Increasing one increases the other (check out "flow coefficient").</p>
<p>Finally, a small centrifugal pump probably won't let you atomize/mist, although that's not a fixed rule. Try positive displacement pumps like diaphragm or gear pumps to get higher pressures at low flows.</p>
| 15372 | Can a small pump be good for atomizing |
2017-05-18T22:20:48.887 | <p>How do I calculate the thrust of ducted contrarotating propellers?</p>
<p>Is there any formula or website calculator or would it have to be done through experimenting with scale replicas?</p>
| |propulsion|thrust| | <p>This is one analysis that gives you information .</p>
<p><a href="https://i.stack.imgur.com/dULrf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dULrf.png" alt="enter image description here"></a></p>
<p>Source : <a href="http://naca.central.cranfield.ac.uk/reports/arc/rm/2218.pdf" rel="nofollow noreferrer">http://naca.central.cranfield.ac.uk/reports/arc/rm/2218.pdf</a></p>
| 15392 | Calculating Thrust for Contra-Rotating Propellers |
2017-05-19T04:08:41.570 | <p>Consider this hypothetical ideal scenario:</p>
<p>If I have a test coupon in a tensile test that has a constant cross-sectional area throughout its length (i.e. along the tensile loading direction) and there are no flaws or imperfections at all, where does necking occur?</p>
<p>In this case stress is uniform throughout the volume of the coupon (no stress concentrations).</p>
<p>Thanks.</p>
| |materials| | <p>If you have a tensile test bar with shoulder fillets, the grips will not be a problem. Then as the metal plastically deforms ( assuming it is ductile) , dislocations will pile-up - work harden. This will eventually produce a local stress concentration > necking. </p>
| 15393 | Tensile Test Necking Location |
2017-05-19T12:48:42.123 | <p>I want to measure the vibration of a roller coaster cart while it is riding down the tracks. Handheld data loggers with a pin on them won't work. I'd need something I can attach or place inside of the roller coaster track. I am worried that conventional data loggers won't be well suited for this because the acceleration of the cart influences the vibration readings when using an accelerometer. I have found exactly 0 results on Google for "measuring vibration of moving objects"-like queries. I want to measure 3-axis G-force.</p>
<p>Is there anything you could recommend me to use?</p>
| |vibration| | <blockquote>
<p>The exact problem is how do I measure vibration in a moving object like a rollercoaster vehicle. Because it accelerates and decelerates the G-forces measured by a vibration sensor using an accelerometer are off since they also include the acceleration of the vehicle. This in contrast to a stationary machine which does not move</p>
</blockquote>
<p>To do this, you do in fact want an accelerometer, and then you want to add to it a thing called a "high pass filter". </p>
<p>The reason this will work is that the "rigid body acceleration" you refer to is at a fairly low frequency compared to the vibration you want to measure. Consider your basic roller coaster ride. Going up the first your accelerometer is going to measure 1g in the vertical direction. Then going down the hill, you measure much less than one g in vertical direction, say 0.2g for a few seconds (this is assuming that your accelerometer always stays in vertical direction and doesn't change with the direction of the car). Then at the bottom of the hill, you are measuring more than 1g, say 2.0g, again for a few seconds. So our signal will have a 1 -> 0.2 -> 2.0 g waveform lasting maybe 5 seconds, or effectively 0.2 Hz. Superimposed on top of that is the vibration you want to measure. That will be at a higher frequency, perhaps 10 - 1000 Hz. So if you apply a high pass filter with a 1 Hz cutoff frequency, you'll get rid of the 0.2 Hz rigid body acceleration and keep the 10 - 1000 Hz vibration. The filtering could be done before the recording is made, or you could just record the whole thing and then filter it out later. </p>
<p>Another entirely different option, would be to use a displacement sensor that measures the distance between the car and the track, which will change very slightly as the car vibrates. These types of sensors can be made with a resolution of better than 0.001", so should be able capture to the vibration of interest.</p>
| 15407 | How do I measure vibration of a moving object? |
2017-05-19T20:28:31.923 | <p>Why does chilled water emerge from the water pipe of air conditioner?</p>
<p>I feel that instead, lukewarm water should come out, the reason being the air conditioner air from the room and also, electricity produces heat energy so as a result warm water should come out. So what is the correct reason for this?</p>
| |electrical-engineering|heat-transfer|heating-systems| | <p>I'm not quite sure of the context of the question or what the water pipeline you are referring to is since most air conditioners that are in homes do not use water, but rather refrigerants. If you are talking about the condensate line from the coil as Chris thought, then yes that is water vapor from the air. Warmer air is able to hold more moisture per volume than cold air and when warm moist air meets the cold heat exchanger (coil) the water vapor in the air will condense on the coil. </p>
<p>As for the sucking out of odors and dust, that is the air return for the air handler that is sucking the air and at the most basic level it is basically pulling air in so that the fan in the air handler has something to blow out. The odors and dust are actually trapped in the filter which is there for the purposes of both filtering out particles for your benefit, but more importantly to protect the coil from all of that crap getting stuck between the fins and making it less effective at heat transfer. </p>
<p>Regarding the electricity, the electricity going into your a/c system is not being deliberately used as a resistive heating element, so the heat generated is only coming off of the motor as a byproduct and is comparatively minimal in magnitude compared to the cooling capacity of the system. </p>
<p>Hopefully that helps and if part of your question is still unanswered, please edit your original question to clarify what it is that you are specifically trying to figure out. </p>
| 15420 | The air conditioner mechanism |
2017-05-20T03:52:53.960 | <p>I often select makeup air handling units for buildings using Greenheck's CAPS tool. When selecting a model, you first select the unit based on your airflow needs. Then it will provide some choices for burners which it gives in units of Btus per hour. It is my understanding that this is a function of the unit to deliver that amount of heat over a period of time based on the volumetric flow that was selected. This would generally follow the rule of thumb for air of approximately: $$\dot{Q}_{Btu/hr}=1.08\cdot\dot{V}_{CFM}\cdot\Delta T_{^{\circ}F}$$</p>
<p>Does this mean that when looking at the burner selection you would have to adjust for a lower air flow rate and subsequently less output or does the temperature delta change accordingly that the heat output for a given burner would still be effectively the same even at the reduced airflow? On any of the software packages, I can't change the flow rate AFTER selecting a unit to see how the numbers change so that I could back into the answer to check for myself and if I simply pick a lower air flow rate, it gives me entirely different burner choices. </p>
| |airflow|heating-systems| | <p>I went back and worked through it, and I was somewhat right. If you request the manufacturer to maximize the burner, you may get a different size burner even when the base unit is the same because typical units are temperature limited. So by reducing the airflow in half, you would double your delta T for a given input, to balance the equations, but the actual selections will not match simply because of manufacturer limits in place to prevent damage to the units.</p>
| 15428 | Makeup Air Gas Burner |
2017-05-21T09:33:55.267 | <p>I would like to determine the massflow rate of (a) steam (b) air flowing through a cylindrical pipe knowing that p_inlet = 5 bars p_outlet = 1 bar. Internal diameter is 0.5m. Length is 30m.</p>
<p>Surface roughness is assumed equal to 5E-05 m. Only steady state situation is necessary. Temperature at the inlet is assumed in both cases to be equal to 150°C. For derivation, one can assume adiabatic flow.</p>
<p>What would be the steps to determine the massflow rate ?
I assume one must first check whether the flow is compressible or not ? Then whether the flow is subsonic, sonic, supersonic ? Then also if laminar or turbulent ?
And then eventually, what would be the equation to be used to determine the massflow rate ?</p>
| |fluid-mechanics| | <p>In pipe flow, we don't typically talk about subsonic / supersonic / transonic flow. We instead differentiate between choked and non-choked flow. Choking occurs when the upstream pressure is greater than the downstream pressure by some factor (dictated by the gas or fluid). That factor is usually ~1.5, depending (roughly) on the ratio of specific heats. You are clearly dealing with choked flow in your case. Choking means the flow speed at the exit is the speed of sound. Any excess pressure past the point of choking does not increase the flow speed. It will merely increase the density of the fluid.</p>
<p>In this case, it does not matter whether your fluid is compressible or gas or liquid phase, because this is a relatively uncomplicated problem. In others, where you may have a restrictor plate or valve, it would. </p>
<p>The equation for the condition of choked flow:</p>
<pre><code>p* / p0 = (2 / (gamma + 1)) ^ (gamma / (gamma - 1))
</code></pre>
<p>where p* is the downstream pressure, p0 is the upstream pressure, and gamma is the ratio of specific heats. If your downstream pressure is lower than p*, then your flow is choked.</p>
<p>In a straight cylindrical pipe we can (almost) always assume flow is laminar, but there are equations to check that for us. The Reynolds number for a pipe flow can be found by:</p>
<pre><code>Re = rho V D / mu
</code></pre>
<p>with rho being density (kg / m3), V being flow speed (m/s), D being the pipe diameter (m), and mu being the dynamic viscosity (N s / m^2). </p>
<p>For choked flow, we can simply use a mass flow rate equation based on pressure differential. If you were looking for mass flow rate without already knowing the downstream pressure, or looking for the pressure drop caused by the friction, we'd have to take account the surface roughness. Similarly more work is required with non-choked flow. </p>
<p>The mass flow rate equation for choked flow:</p>
<pre><code>mdot = Cd A sqrt(gamma rho0 p0 (2 / (gamma +1)) ^ (gamma + 1) / (gamma - 1))
</code></pre>
<p>with:</p>
<p>Cd - the coefficient of discharge, defined next</p>
<p>A - pipe exit area</p>
<p>gamma - ratio of specific heats</p>
<p>rho0 - upstream density (which you should compute based on T, p0)</p>
<p>p0 - upstream pressure</p>
<pre><code>Cd = mdot / (A sqrt(2 rho deltaP))
</code></pre>
<p>with deltaP being the pressure drop across the exit (here, 5 bars - 1 bar = 4 bars). You'll need to convert units here as appropriate (i.e. to Pascals).</p>
<p>The coefficient of discharge basically is a refinement to the mass flow rate; it takes into account that the cross section of the flow in the orifice at its maximum speed is not the same size as the total area of the orifice. If you're bored and want to learn more about that phenomena, look up vena contracta. </p>
<p>There's a wide array of pipe flow, valve, orifice, etc. problems out there. If you're looking for more information, Crane's TP-410 publication is the de facto reference for these sort of standard pipe flow problems. I also really like Robert Blevin's Applied Fluid Dynamics Handbook, as it has a host of information not available anywhere on the internet (for estimating pressure losses due to all sorts of pipe configurations). </p>
| 15453 | Mass flow rate (air and steam) through a cylindrical pipe |
2017-05-21T20:21:28.723 | <p>I know what pattern draft is and it's usefulness in removing castings from rigid moulds. That's the definition you can find anywhere. </p>
<p>My problem is when you sand cast. Sand casting is the most common amateur metal casting technique, with lots of YouTube examples. Yet all the decent examples have a draft incorporated. Since you destroy the sand mould when extracting the casting, I can't see any benefit from a draft. It just adds an unnecessary level of complexity when designing and making the pattern. Any shape could be successfully removed, so what's the point of pattern draft in sand casting?</p>
| |casting| | <p>You need pattern draft to be able to remove the pattern from the sand before pouring. If you don't have any / sufficient draft then edges can be weakened or damaged / broken when you remove the pattern from the sand or plaster - whatever material it is. It is not to help remove the casting as the sand mould is destroyed.</p>
| 15455 | What's the point of draft in sand casting |
2017-05-22T01:17:33.503 | <p>I've designed a whole lot of different parts and products in my time: Plastic, metal, lighting products, electrical products and automotive parts. I generally can look at a product and figure out why did they do that. Here's one that has me baffled. </p>
<p>Its the bottom of a consumer grade vacuum cleaner. Molded in the plastic is a slot (which I'd generally refer to as a bayonet mount). I can see that the mount definitely is out of normal die draw (large undercut) and requires a slide (and that tooling is definitely not cheap...) I can't figure out what that mount is for. It doesn't seem to match the cord or any of the accessories that I can see. </p>
<p><a href="https://i.stack.imgur.com/bJGX0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bJGX0.jpg" alt="Vacuum cleaner mystery mount"></a></p>
<p>Any ideas on what that 'bayonet mount' is for? Why would the manufacturer spend real investment dollars for tooling something so obtuse? </p>
<p>Possible </p>
<ul>
<li>Something they use in their manufacturing process? </li>
<li>Something to ensure the part control during the molding ejection process from the die? Note: I've seen molded parts get stuck in the wrong half of the tool and that can be quite ugly... The other side of the part is heavily engineered, with lots of ribs and other details.</li>
<li>Could it be for planting a sales/marketing sign when the product is sitting in a sales display? </li>
</ul>
<p>Any ideas? </p>
| |plastic|molding|injection| | <p>This was an unexpected surprise. The slot is for a mating element on the lower tube extension. I totally missed this. When stowed this way:</p>
<ul>
<li>the storage footprint is small and stable</li>
<li>you can easily move the entire assembly in and out of a storage closet with one hand (by grabbing the handle on the base unit.)</li>
</ul>
<p>Thanks to Solar Mike and Zimano for their responses. Separate answer posted here to enable a photograph of the assembly.</p>
<p><a href="https://i.stack.imgur.com/MgMm8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MgMm8.jpg" alt="vacuum cleaner hose storage slot"></a></p>
| 15459 | Plastic Vacuum Cleaner Part -- Why did they design it this way? |
2017-05-22T23:45:09.687 | <p>How much graphene is required to coat a 5 inch (diameter) sphere if the graphene layer must be 35 nanometers thick?</p>
| |materials|mathematics|geometry| | <p>The volume $V$ needed to cover the surface is equal to the spherical shell volume with $d_{\text{inner}}=5 \text{ inch}=0.127\text{ m}$ and $d_{\text{outer}}=d_{\text{inner}}+2\cdot 35 \cdot 10^{-9}\text{ m}$. Remember to use everything in SI-units.</p>
<p>$$V=\frac{4}{3}\pi\left[\left(\frac{d_{\text{outer}}}{2}\right)^3-\left(\frac{d_{\text{inner}}}{2}\right)^3\right]\approx1.7734\cdot 10^{-9} \text{ m}^3$$</p>
<p>If the density of graphene is given by $\rho$, then the necessary mass $m$ of graphene is given by $m=\rho V$.</p>
<p>Note, that the answer is very similar to the result obtained by Angus Murray, as the outer diameter is almost the same as the inner diameter. But I would not use approximative formulas as long as there is an exact formula because it will not be necessary to estimate the error that you have introduced by using this approximation.</p>
| 15472 | Graphene distribution over a sphere |
2017-05-23T03:11:25.613 | <p>How can permeation units in Darcy be converted to the SI form. For example, how would 9.0 x 10-6 darcy be converted to SI units of permeation? How would it also be converted to the permeation units: (cm^3 . mm)/m^2 . d . atm ?</p>
| |measurements|mathematics|unit| | <p>For your question, 1 m2 = 10^12 Darcy.</p>
<p>For the next question, I think your question is units involved in Darcy's Law, i.e., the Darcy's units and SI units.</p>
<p>In general, we define the permeability of porous media as 1 Darcy, it means the porous media can transmit 1 cm3/s of water with viscosity of 1 cP (1 mPas) under pressure gradient of 1 atm/cm cross an area of 1 cm2. The units involved here are called Darcy's units.</p>
<p>For SI unit, we use flow rate of m3/s, viscosity of Pas, pressure of Pa, length of m, cross section area of m2, and we can get the permeability unit with m2.</p>
<p>If we use the units system for both cases, Darcy's equation is the same without any conversion factor.</p>
| 15473 | Darcy to SI permeation units |
2017-05-24T07:14:55.997 | <p>I am currently doing a control systems course and I was lead to believe that if a pole lies on the imaginary axis in the s-plane (i.e s=0), then the system was marginally/critically stable. </p>
<p>However, I come across this transfer function and plotted it in Simulink. </p>
<p><a href="https://i.stack.imgur.com/MgLQF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MgLQF.png" alt="enter image description here"></a></p>
<p>Analysis shows that there are 3 poles at s=-1, s=-3 and s=0. So because there is a pole at s=0, the system should be marginally stable right? But the output of the transfer function shows this:</p>
<p><a href="https://i.stack.imgur.com/d2Bpf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d2Bpf.png" alt="enter image description here"></a></p>
<p>To me, that is showing the system is unstable because the output is getting bigger and bigger as time goes on. Is this correct? Or is my understanding still flawed? </p>
<p>If the system is unstable, then why aren't any of the poles in the right half of the s-plane? </p>
<p>Thanks</p>
| |control-engineering|stability| | <p>Your system is open loop stable as the poles are at $s=-1$, $s=-3$ and $s=0$. Note, that if the order of the pole at $s=0$ is greater then 1, then the open loop system is also unstable.</p>
<p>But closing the loop changes the poles of the system. If $F(s)$ is your transfer function of the open loop system, then the transfer function of the closed loop system is:</p>
<p>$$G(s)=\frac{F(s)}{1+F(s)}=\frac{21}{s^3+4s^2+3s+21}.$$</p>
<p>The <a href="http://www.wolframalpha.com/input/?i=s%5E3%2B4*s%5E2%2B3*s%2B21%3D0" rel="nofollow noreferrer">poles of $G(s)$</a> are $s \approx -4.4022$, $s \approx 0.2011 - 2.1748 i$ and $s \approx 0.2011 + 2.1748 i$. Hence, the closed loop system is unstable.</p>
| 15485 | Control Systems: Why is this system unstable? |
2017-05-24T15:24:00.093 | <p>What is the logic behind using a wall tap/piezometer to measure the static pressure. Is it that near the walls, the fluid velocity is zero, because of the no slip condition and hence the pressure measurement near the wall would only be static pressure. There is this point in the book that i am consulting, it says "there is no pressure variation normal to straight streamlines, this fact makes it possible to measure static pressure in a flowing fluid using a wall pressure tap". Does this point mean that because there is non-variance of pressure in normal direction to the straight streamlines, thus giving a single value for static pressure, whereas in regions of curved streamlines, we would have progressively increasing values as we go away from the center of curvature of the family of curved streamlines. Also, why do we require a separate static pressure measuring instrument alongside a pitot tube (for fluids passing through a pipe) in order to measure velocity through the Bernoulli eq. Why can't we calculate velocity directly as is done, for free streams of fluids, such as river water. </p>
| |mechanical-engineering|fluid-mechanics|pressure| | <p>A piezometer is a device which senses pressure by producing electric signals proportional to the applied pressure. It is attached on the walls of the pipe so that the fluid pressure acts normal to it. Since the fluid flow is parallel to the walls, the fluid flow is neither obstructed by the piezometer, nor does it experience any momentum transfer. The sole force acting on the piezometer will be that caused by the hydrostatic pressure. So the need for measuring velocity of flow does not arise. This has nothing to do with no slip condition. We can place the piezometer at any point in the direction of streamline. It is more convenient to fix it on the walls.</p>
<p>In case of the type of gauges mentioned in your second question, they are placed normal to the flow so that they experience both hydrostatic pressure force and force due to momentum transfer(since they obstruct the flow). So we need to subtract the hydrostatic pressure to find the flow velocity.</p>
| 15494 | Measurement of static pressure |
2017-05-25T10:46:20.190 | <p>For a mechanical design, I would need a small/light axle, having high precision (as less loose as possible and reduced drag). With precision, there is not a very strict definition, however, I target ~0.2º of radial rotation (axial movement is not that an issue).</p>
<p>The length of the axle is around 24mm, on which I would attach a few elements (~50g).</p>
<hr>
<p>The first idea is to use a bearing on each side of the axle, but I am not sure what is the good practice for attaching the bearing.</p>
<p>Following is my current sketch:</p>
<p><a href="https://i.stack.imgur.com/snj03.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/snj03.png" alt="enter image description here"></a></p>
<p>The closest example of what I am doing is the axle in the actuator of a Hard-Disk, however, the bearing there is internally tapped for fixation. Most of the bearing I found on the market are not tapped. My feeling is that tapping the bearing is not a good idea.</p>
<p><a href="https://i.stack.imgur.com/kjyoV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kjyoV.jpg" alt="enter image description here"></a></p>
<hr>
<p>My question is:</p>
<p><strong>What are the commonly use pieces, practices, terms and fixations for this type of problem?</strong></p>
<p>I understand is a quite broad question, but either I miss terminologies and in any case I did not found answers online. Also, I do not need a deep answer, but some indications of where to look further.</p>
<p>Others possible tags: axle, matched-set</p>
| |mechanical-engineering| | <p>Axle type rod is available under different names on-line. Try searching for "8mm linear shaft", "linear rail shaft", "ground polished shaft" just to start.</p>
<p>Bearings that fit snug don't require much of a fastener to keep them from walking off of the shaft. You could machine a hole through the end, or a groove around the end, and use an R-clip or c-clip snapped onto the end. It would probably be overkill, but you could drill and tap a hole straight into the center of each end for a screw and washer.</p>
| 15506 | How to design a precision (small) axle? |
2017-05-25T16:50:27.427 | <p>Is there a formula for the burst working pressure of a spherical pressure vessel that takes into account its material peoperties? It should be noted that this is not a concentric sphere design (it has only one radius) .This is a formula for a pipe: <a href="http://www.engineersedge.com/calculators/pipe_bust_calc.htm" rel="nofollow noreferrer">http://www.engineersedge.com/calculators/pipe_bust_calc.htm</a> </p>
| |pressure|statics|stresses|pressure-vessel| | <p>For spheres,stresses in the material is same in all directions.So ,hoop stress and longitudinal stresses are the same.
Using similar abbreviations, as P for pressure(gauge) inside sphere ,FS for factor of safety,S for allowable stress and additionly
Ri for inner radius,Ro for outer radius
P(gauge pressure)=((Ro^2-Ri^2)×S)÷((Ri^2)×FS)
For calculation of brusting pressure,take S as ultimate
stress for a given material and put FS=1</p>
| 15514 | Burst pressure of a sphere |
2017-05-26T11:06:28.477 | <p>I was curious about this. If i have a tube which has water flowing in it. and mounted vertically a small extension tube with a loop. The water will go horizantally and only a few drops will go vertically. Why is that? and how can u determine how much water will go vertically? I tried to search for a scientic explaination but didn't find anything. What if I want to design a tube with this loop but I don't want any water to go vertically?</p>
<p>Here is a diagram if you don't understand: <a href="https://i.stack.imgur.com/M6ofE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M6ofE.png" alt="enter image description here"></a></p>
<p>Is there anyway to prevent the water from going verically if the tube is blocked? </p>
| |mechanical-engineering|fluid-mechanics| | <p>Water likes to take the path of least resistance.</p>
<p>In this scenario, you have 2 paths for the water to take when it reaches your T-junction.</p>
<p>If we compare the two paths, we can see why it doesn't want to travel through the vertical tube. First, to go through the vertical tube it would need to overcome gravity. The tube seems thinner, so it would also have greater frictional losses as it travels through it. To add to that, it has a bend, which is another source of friction loss compared to a straight pipe.</p>
<p>If you're familiar with basic DC electric circuits, you may know that if you have current travelling through parallel resistors, the higher current will flow through the lower resistance branch. This is an analogous situation. The current is like the flow rate, the resistance is like the frictional and gravity losses, and the voltage is like the pressure.</p>
<p>You can use the equations of <a href="https://en.wikipedia.org/wiki/Bernoulli%27s_principle" rel="nofollow noreferrer">Bernoulli's Principle</a> to get some idea on how these losses will effect the flowrate.</p>
<p>If you want to prevent water flow vertically when the tube is blocked, you need to make sure the pressure losses in the vertical tube due to gravity and friction are greater than the maximum pressure you can build up. This will depend on how you are getting your flow, if it's a pump, the pump may stall or throttle itself once it reaches max operating pressure. If it's flowing from a higher point, the safe thing to do would be to make sure your tube is higher than the high-point you feed in from. (you can probably get away with slightly below the high point due to friction losses on the way there, especially depending on how long the pipes are).</p>
| 15531 | Why would the flow go horizantally not vertically in the tube? |
2017-05-27T08:46:39.313 | <p>To save on material, I am requested to design a continuously supported beam with a maximum bending capacity that is half of its maximum bending moment and to strengthen the beam at certain intervals where its capacity is exceeded. What is a feasible strengthening scheme without changing the beam's overall dimensions significantly.</p>
<p>This beam is going to be used as a waling where stiffeners will be added at strut-waling connection. This is also the location where moment capacity will be exceeded.</p>
| |structural-engineering|civil-engineering|steel| | <p>Castellated beam. Cut the web with torch, and re-weld web at higher flange distance.</p>
| 15547 | How to strengthen an I beam |
2017-05-27T17:35:34.510 | <p>I have a beam. I have its Young's modulus. I don't have its maximum acceptable deflection. How can I determine the maximum acceptable deflection without breaking the beam? </p>
| |beam| | <p>Yield strength of steels is generally defined by 0.2% deflection . So if you do not exceed 0.2% strain under load you will not permanently deform it .The specification of the steel will give the minimum Yield Strength. Then if your load does not cause a stress greater than this specified yield strength it will not "break". </p>
| 15551 | How to find maximum acceptable deflection? |
2017-05-28T04:34:07.040 | <p>I recently bumbped into a rather basic but interesting question on aeroelasticity. I've learned to derive the fluttering critical speed from Pines's theory but it involves some spring stiffness like for normal spring $K_h$ and torsion spring $K_t$.</p>
<p>However, I want to see if I can derive the same thing if the wing is now treated as a continuous beam-rod model (i.e. torsion and bending). The equations of motion are: </p>
<p>$ \frac{\partial^2}{\partial y^2} \left( EI \frac{\partial^2 h}{\partial y^2} \right) + m \frac{\partial^2 h}{\partial t^2} + m x_\alpha \frac{\partial^2 \alpha}{\partial t^2}+L=0$</p>
<p>$ -\frac{\partial}{\partial y} \left( GJ \frac{\partial \alpha}{\partial y} \right) + I_\alpha \frac{\partial^2 \alpha}{\partial t^2} + mx_\alpha \frac{\partial^2 h}{\partial t^2} - M = 0$</p>
<p>For simplicity, let $h(y,t)=0$ and $\alpha=s(y)e^{pt}$ so we can focus entirely on the torsion dynamic response. Let $L=qca_0(\alpha+\alpha_0)$ and $M=qcea_0(\alpha+\alpha_0)$. But now things get out of my control as I don't know how to solve for $p$ (to let its real part positive), given $s(y)$ unsolved too. </p>
| |beam|mathematics|aerodynamics| | <p>By writing $\alpha=s(y) e^{pt}$, you've assumed that the solution is separable, i.e. that it can be separated into a part that depends only on y and another part that depends only on t. This is pretty common for this type of equation and it should work. The typical next step here would be to separate the variables in the equation. i.e. move everything that depends on s(y) (and its derivatives) to one side, and everything that depends on $e^{pt}$ (and its derivatives) to the other side. Then, since the left depends only on y, and the right only on t, the only way this can work if is both sides are equal to the same constant. You then just need to find that constant. Typically you'll proceed by assuming something like $s(y) = A \cos(kpy)+B \sin(kpy)$, where A and B are unknown constants and k is the wavenumber. Then just plug in and solve. There will be an infinite number of solutions to the equation. Each solution is a combination of k, A, B that solve the equation (and the boundary conditions), and represents an eigenmode of vibration. Typically, you'll keep only the first few (or maybe only first one) eigenmodes, and then find the time response of the modes. </p>
<p>This page might help, <a href="http://www1.aucegypt.edu/faculty/mharafa/MENG%20475/Continuous%20Systems%20Fall%202010.pdf" rel="nofollow noreferrer">http://www1.aucegypt.edu/faculty/mharafa/MENG%20475/Continuous%20Systems%20Fall%202010.pdf</a> , or pick up any textbook on "vibrations of continuous systems".</p>
| 15556 | Wing as cantilever beam (beam-rod) model for aeroelasticity fluttering analysis |
2017-05-28T22:14:22.737 | <p>By this I mean can any DC to AC variable frequency drives invert digitally (no moving parts) and if so how do these work? Is this possible for both sine and square wave inverters?</p>
| |electrical-engineering|frequency-response|ac| | <p>This question is rather confused, but I'll take it as <i>"Can AC power be produced from DC power without mechanically moving parts"</i>.</p>
<p><b>Of course</b>. That's what devices called "inverters" do. Most inverters are designed to work at a single frequency, but they certainly don't have to be.</p>
<p>Think of class D amplifiers as examples. Clearly they can produce a wide range of frequencies. In the case of amplifiers, the frequencies they put out are the frequencies in a small control signal. However, that control signal could just as well be produced internally if you have a way of describing it, like a sine of a particular frequency.</p>
<p>There is not much difference between a class D amplifier and a modern sine wave inverter (except for resonant design, beyond the scope of this answer). In the case of the inverter, the control signal is generated internally, but the power output electronics are quite similar.</p>
<p>Another obvious existence proof of such things are uninterruptable power supplies (UPSs). When the input power goes away, the power source is the internal battery. That is inherently DC, yet the output power is AC at a regulated frequency and amplitude.</p>
<p>In ye olde days, this kind of conversion was done with motor-generators. There is still a niche for those today, although that niche is small and getting smaller as power electronics advance.</p>
<p>Even utility-scale DC to AC conversion is done with solid state electronics nowadays. For example, here is a picture from Google Earth of such a plant in Ayer Massachusetts:</p>
<p><a href="https://i.stack.imgur.com/nGVlS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nGVlS.jpg" alt=""></a></p>
<p>DC power lines from dams nearly 1000 miles away in northern Quebec come in at top left. The big square site at lower right converts that to AC at 60 Hz, and feeds it onto the large AC power lines shown horizontally at the bottom.</p>
<p>Transforming that much power at such high voltages and currents takes a large facility, but it is ultimately done with semiconductors.</p>
| 15569 | Are there any variable frequency drives that work digitally? |
2017-05-29T01:15:01.803 | <p>I saw crane on very tall building. How do they put such heavy item on such high building? </p>
<p><a href="https://i.stack.imgur.com/4bd3P.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/4bd3P.jpg" alt="enter image description here"></a></p>
| |civil-engineering|construction-management| | <p>They lift it up in pieces and assemble the crane in place using a second mobile crane they bring in to build it.</p>
<p>The heaviest parts will be the winch assembly and the counter weights.</p>
| 15573 | How they put heavy cranes on top of the building? |
2017-05-29T14:26:28.013 | <p>In the <strong>real world</strong>, would this pinned frame exhibit a deflection in the fixed, left hand side column?</p>
<p>Arguments from peers largely surround:</p>
<p>1) There <strong>wouldn't</strong> be left-column horizontal deflection because the moment cannot apply through the pin to the column. </p>
<p>2) There <strong>would</strong> be left-column horizontal deflection because although the moment doesn't travel through the left-column pin, in reality the right side will sag and pull the left-column towards the right.</p>
<p><a href="https://i.stack.imgur.com/xjYd0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xjYd0.png" alt="enter image description here"></a></p>
<p>Exceptional answers will explain if points 1) & 2) are correct; provides additional information we haven't considered; and the inclusion of visual diagrams if relevant. </p>
| |structural-engineering|deflection| | <p><a href="https://engineering.stackexchange.com/a/15584/255">mg4w's answer</a> has some great examples of how your modelling assumptions aren't true in reality. But even ignoring all of those there will <em>still</em> be deflection in the left hand column.</p>
<p>You may not find any deflection in your analysis, see for example <a href="https://engineering.stackexchange.com/a/16378/255">Sam's answer</a>. <em>But this is because you're doing an over-simplistic analysis</em>.</p>
<p>We often assume that we have first-order linear static behaviour. This is because it is easier/faster and normally gives sufficiently useful results. However, a second-order geometrically non-linear (GNL) analysis is closer to reality.</p>
<p>If you hold a piece of string, with a weight applied in the middle, you will find that the ends of the string are being pulled inwards. This is because string has very low bending stiffness, and instead axial stiffness governs. With the top beam in your example it is likely that bending stiffness is significantly greater than axial stiffness, but there will still be <em>some</em> component of axial load. GNL analysis will pick this up. The axial force in the beam will apply a horizontal force on to the top of both columns; this horizontal force will cause the columns to deflect inwards.</p>
| 15583 | Structural Frame Real World Deflection |
2017-05-30T21:40:41.113 | <p>Our small radio-control model cars have shims, spacers and washers. I am tasked with clarifying the manual and getting parts naming consistent. </p>
<p>We have these items from less than .5mm thin up to 1/8" thick. I'm thinking of calling everything under .5mm a shim, and everything .5mm and thicker that is available off the shelf a washer. And spacers will be anything .5mm and thicker that we have manufactured ourselves for specific needs on the car. </p>
<p>How do engineers define the difference between these terms?</p>
| |mechanical-engineering|computer-hardware| | <p>No real difference between a shim and a spacer. A shim is used to adjust the space/clearance between faying surfaces (where two pieces connect). Generally as a means of getting away from the cost of tight tolerances. Sometimes as a wear part that gets replaced to hold a particular clearance (e.g. the selectable shims that were used to adjust the clearance in my overhead cam on my old motorcycle). A washer is used to distribute the load from the head of a bolt or screw (or to keep your faucet from leaking). Also may perform a locking function. Nothing to keep you from using a washer, or stack of washers, as a shim - to adjust a clearance. I'd label the part by function. If you're using it to adjust a clearance call it a spacer. If you're using it under the head of a screw or bolt, call it a washer. In my opinion the function of the part in its application is more useful and relevant than an arbitrary name.</p>
| 15608 | How do we specify among a shim, spacer, and washer? |
2017-05-31T08:01:01.820 | <p><a href="https://i.stack.imgur.com/zYDFY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zYDFY.png" alt="system" /></a></p>
<p>I need to model the system in figure, where turbine, shaft1 and gear1 are a unique rigid body, while shaft2 can be modeled like a torsional spring. Using as coordinates Theta1, clockwise angular displacement of the turbine, and Theta2, anticlockwise displacement of the generator, i obtain these equations of motion</p>
<p><a href="https://i.stack.imgur.com/Vpnje.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vpnje.gif" alt="differential equation" /></a></p>
<p>where tau is the transmission ratio of the gears and J0 the equivalent moment of inertia of the turbine-shaft-gears block.</p>
<p>The stiffness matrix is not symmetric and it should be impossible, where am I wrong?</p>
| |mechanical-engineering|stiffness| | <p>One way to get this right is to start by ignoring the gearbox and modelling the system with three DOFs, i.e. the angular position of Shaft 1, and the positions of the two ends of Shaft 2. That gives the equations of motion as
$$\begin{bmatrix}J_0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & J_1 \end{bmatrix}
\begin{bmatrix} \ddot\theta_1 \\ \ddot\theta_g \\ \ddot\theta_2 \end{bmatrix} +
\begin{bmatrix} 0 & 0 & 0 \\ 0 & k_{t2} & -k_{t2} \\ 0 & -k_{t2} & k_{t2} \end{bmatrix}
\begin{bmatrix} \theta_1 \\ \theta_g \\ \theta_2 \end{bmatrix} = 0$$
or $$M_{3\times3}\,\ddot\theta_{3\times3} + K_{3\times3}\,\theta_{3\times3} = 0.$$</p>
<p>Then, apply the gearbox constraint, i.e. $\theta_g = -\tau\theta_1$. That is equivalent to the transformation matrix
$$\begin{bmatrix} \theta_1 \\ \theta_g \\ \theta_2 \end{bmatrix} =
\begin{bmatrix} 1 & 0 \\ -\tau & 0 \\ 0 & 1 \end{bmatrix}
\begin{bmatrix} \theta_1 \\ \theta_2 \end{bmatrix}$$
or
$$\theta_{3\times3} = L\,\theta_{2\times2}.$$</p>
<p>So the $2\times2$ stiffness and mass matrices are
$$\begin{align}
M_{2\times2} &= L^T M_{3\times 3}\, L \\
K_{2\times2} &= L^T K_{3\times 3}\, L
\end{align}$$
which are symmetric.</p>
<p>Following a systematic procedure might seem more longer than trying to write the answer down "by inspection," but it's more likely to avoid mistakes. </p>
<p>For example, one of the diagonal terms in the <em>correct</em> stiffness matrix is $\tau^2 k_{t2}$, which doesn't appear anywhere in the OP's equations of motion. That suggests the OP's mistake was getting confused about the units being used to measure the various torques and rotations in the system in terms of $\theta_1$, $\theta_2$, and $\tau$.</p>
| 15612 | Asymettric stiffness matrix for a generator-gearbox-turbine system |
2017-06-01T01:05:22.337 | <p>Basic Googling on the mechanical generation of oscillating motion (moving something back and forth in the manner of a windshield wiper or a flapping wing) from a source of circular motion comes up with a concept known as a "four-bar linkage" or "crank rocker". For example, this video shows very clearly a simple way to create a windshield-wiper motion:</p>
<ul>
<li><a href="https://youtu.be/4tIo3AQQiU8" rel="nofollow noreferrer">Crank-Rocker Four-Bar Linkage</a></li>
</ul>
<p>The problem is, <em>this doesn't look anything like a windshield wiper.</em> It has a bar attached to the tip of the "wiper" pulling it back and forth along its arc. But in an actual windshield wiper (or wing, for that matter) the moving force comes from the <em>base,</em> not the tip.</p>
<p>So how do you go about generating oscillating motion from the base of the oscillating device, windshield-wiper style? Assume you're starting with a motor providing circular motion, as in the video.</p>
| |motors| | <p>The wiper arm is on a shaft, which passes through the car's bodywork. What you don't see is another arm (perhaps only an inch or two in length) on the backside of the shaft, with the crank rocker arrangement attached to it.</p>
| 15625 | How does a windshield wiper work? |
2017-06-01T03:16:49.607 | <p>This is a follow on from my previous question, <a href="https://engineering.stackexchange.com/questions/13471/will-standard-m5-t-slot-nuts-fit-these-different-2020-aluminium-extrusions">Will standard M5 T-slot nuts fit these different 2020 aluminium extrusions?</a></p>
<p>The extrusion has this profile:</p>
<p><a href="https://i.stack.imgur.com/MoVBE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MoVBE.jpg" alt="2020 Aluminium Extrusion BKK#1"></a></p>
<p>and here is a more in-focus photo, of four, in a 2 x 2 configuration:</p>
<p><a href="https://i.stack.imgur.com/Vzce3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Vzce3.jpg" alt="2020 Aluminium Extrusion BKK#2"></a></p>
<p>The gap, between the bottom of the flange and the top of the circle, is about 4.4 mm high. The M5 hexagonal nuts are 3.85 mm thick, and 7.85 mm wide.</p>
<p>I have been using these 2020 extrusions with regular M5 hexagonal nuts (and M5 bolts, obviously). </p>
<p>However, I have recently found a need to use M3 bolts as well (to secure PCB mounted mechanical switches and endstop holders to the extrusion, for a 3D printer, see <a href="https://3dprinting.stackexchange.com/questions/4173/complete-list-of-fasteners-for-the-sintron-kossel">Complete list of fasteners for the Sintron Kossel</a>).</p>
<p>The width of the slot is 5.4 mm and regular hexagonal M3 nuts are also around the same width (5.38 mm), and hence slip out of the slot. I guess that I <em>could</em> use a wider washer to hold the nut in place, but does anyone know of a "fat" M3 nut that is available, preferably one that is the same size as an M5 nut, but with a smaller M3 bore?</p>
<p>The problem has arisen because I should be using "European" 2020 aluminium, which takes T-nuts, and those come in a range of bores (M3-M6). However, I have been unable to obtain any, in Bangkok.</p>
| |aluminum|fasteners| | <p>Another option would be to make your own t-nut.</p>
<p>If you have a mill/CNC, here are some instructions to <a href="http://rick.sparber.org/tn.pdf" rel="nofollow noreferrer">Make a T-Nut</a></p>
<p>If you don't have access to a mill, you could make a pretty good one out of some 4mm x 5mm <a href="https://www.mcmaster.com/#standard-aluminum-sheets" rel="nofollow noreferrer">bar stock</a> and tap your own holes. All you would need is a <a href="http://rads.stackoverflow.com/amzn/click/B00HQONFVE" rel="nofollow noreferrer">drill press</a> and an <a href="http://rads.stackoverflow.com/amzn/click/B01FTFX40S" rel="nofollow noreferrer">M3 Tap</a>. The nut may not be as rigid as a t-nut that precisely matches you channel and you can't put as much torque on 4mm of bar stock aluminum as you could on a steel t-nut; but, for most application it would probably be sufficient.</p>
| 15628 | Fat hexagonal M3 nuts, with outer diameter greater than 6 mm |
2017-06-02T10:53:44.943 | <p>i am first year engineer and this is my first class on electronics.
Can please someone explain why ix' = 0.2A ? <img src="https://i.stack.imgur.com/kMlA1.jpg" alt=""></p>
<p>** I mean it is 0.2 if we add the resistors, but wouldn't it be 0.5 after the 6Ω? </p>
| |electrical-engineering| | <p>In that circuit the resistors are in series, so the current through them is the same. In addition, to calculate the current, you can sum the values of the resistors (because they are in series) and then apply Ohm's law: $I=\frac{V}{R}=\frac{3V}{(9+6)\Omega}=0.2A$</p>
| 15645 | Current calculation |
2017-06-04T02:50:38.047 | <p>So I tried equating the sum of forces in the x,y and torque about $C$ to zero and found $B_x$ and $C_x$.</p>
<p>But how to find $B_y$ and $C_y$? I have 3 equations and 4 unknowns? </p>
<p><img src="https://i.stack.imgur.com/gxu1T.png" alt=""></p>
| |structural-engineering|civil-engineering| | <p>Creating an FBD in joint A:<a href="https://i.stack.imgur.com/ccnCU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ccnCU.png" alt="enter image description here"></a></p>
<p>We can derive that member AC is under compression. <strong>Note: you don't need to solve for the reactions since we can answer directly using method of joints</strong></p>
<p>Euler's buckling formula states that:</p>
<p><a href="https://i.stack.imgur.com/jGwXU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jGwXU.png" alt="enter image description here"></a></p>
<p>Therefore, if you check your column there, the critical buckling load is equal to 37.285 kN (Compression). <strong>Note: use K = 1.0 since both ends are pin-connected</strong></p>
<p>So to answer the first question: <strong>NO, it is not strong enough</strong></p>
<p>the second question reverses the function mentioned above, given capacity of AC = 37.285 kN, we can get AB = 74.570 kN, therefore, <strong>the point load capacity at joint A is equal to 64.580 kN.</strong></p>
| 15663 | Euler buckling formula,triangular structure |
2017-06-04T04:40:54.390 | <p>In the making of an inflatable toy ball (think of a basketball), a check valve is used to assist the user in the inflation of the ball by preventing backflow of inserted air. Is this check valve made of any metalic parts (such as conventional check valves) or is it all plastic? Also, what type of valve is it? Is there a maximum pressure this type of check valve can sustain before becoming damaged?</p>
| |pressure|valves|pressure-vessel| | <p>It depends on valve design and the application needs.Can ben matal or teflon lined,like <a href="http://www.ceramic-valves.com/96-sight-glass-ball-check-valve.html" rel="nofollow noreferrer">PFA lined ball check valve</a> for corrsive applications.</p>
| 15664 | Toy ball check valve functionality and details |
2017-06-04T17:42:56.103 | <p>Suppose that I want to transport gases in a tube in only one direction. Can a liquid check valve be used in place of a gas check valve or is there a specific reason to use only a gas check valve?</p>
| |pressure|valves|pressure-vessel| | <p>Industrial valves are ( or were ) marked W O G . Meaning they are suitable for water, oil, and gas, ( not gasoline .). It is up to the user to determine if the valve will function in his particular application. </p>
| 15671 | Liquid vs. gas check valves |
2017-06-05T00:48:41.710 | <p>Recently, I heard that there is a law/ OSHA/ ASME piece of regulation dictating that hydrogen should be transferred (suppose from a tank of hydrogen to another location) by using specially approved tubing. Is this true?</p>
| |pressure|safety|compressed-air|pressure-vessel| | <p>Well, I have used regular acrylic tubing for the gaseous state. However, if any spark contact may occur, use spark resistors, a bubbler, ect. If it is in liquid state I would assume that more would be required. My project this year used hydrogen as a fuel source produced through electrolysis that was run through solar energy. My hydrogen output was in a gaseous state for efficiency. </p>
| 15674 | Tubing for explosive gas or hydrogen transfer |
2017-06-05T02:19:39.920 | <p>Damascus steel is a multilayered hand-forged steel used in the past for arabic and japanese weapons and now for kitchen knives and various connoisseur blades. Usually, two alloys are used alternatively to stack layers having different properties and shades.</p>
<p>I wonder if it still has a mechanical superiority over modern carbon steels, especially with <a href="https://www.crucible.com/eselector/grades&names/0%7Egrades&namestitle.html" rel="noreferrer">"super steels" and tools steels</a> like CPM 3V. By technical superiority, I mean better properties regarding fatigue resistance, impact resistance, tear propagation, edge durability, corrosion resistance, ultimate average tensile strength and so on.</p>
<p>Are there any studies about that ?</p>
| |mechanical-engineering|steel|alloys| | <p>Wootz steel/ Damascus steel is an incredibly beautifully made alloying process and for this alone yes it is still relevant. I personally am working on all sorts of variables in the process of making said alloys.
It has many uses, perhaps not in the industrial scene, but yes it has its place.
Now, on to the "super steels" yes there are super steels that are stronger, hold a keener edge for longer etc etc etc, however as a knife maker I enjoy making wootz/damascus steels. And I am just guessing that you do not know how much some of these super steels cost?
Besides, as I forge my own blades, I am not one of these guys that just removes material and considers themself a knife maker! I have a healthy business, so yes, I believe damascus has its place and will continue to for a long long time. </p>
| 15676 | Steel : is Damascus steel still relevant compared to modern alloys? |
2017-06-05T20:12:26.233 | <p>I have a building that is 7 story height (which 5 stories are basement floors which means they are underground) My water level is at 1.5 m from the surface. the stories have typical rectangular shape of 30 m x 36.2 m with a grid of columns as shown in the picture below.
When modeling the raft foundation that is subjected to gravity loads and water pressure (opposite to gravity loads) and since the raft is at a level of 19.5 m from the ground surface it is subjected to a lot of uplift the overall gravity load is of 108858 KN as for the uplift is (19.5-1.5)*10*30*36.2 = 195480 KN which is greater than the gravity loads so their will be an uplift of the building.</p>
<p>How to solve this problem ? I mean how many solutions that are available today for this problem and what is the most economical one ?</p>
<p><a href="https://i.stack.imgur.com/qOKY6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qOKY6.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/I1AE1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I1AE1.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/ajdd4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ajdd4.png" alt="enter image description here"></a></p>
| |structural-engineering|civil-engineering|structural-analysis|foundations| | <p>In such a case following multiple solutions are suggested.</p>
<ol>
<li>Tension piles are the best solution. Raft often creates uplift pressure if the water table is high or may vary season to season.</li>
<li>Now the cheapest one: Place pipes at various and multiple positions in the raft vertically to release the pressure. These pipes shall be interconnected above the raft in an appropriate slope towards a harvesting tank. Water will rise in the pipe which can be dropped in the tank from where can be pumped out for gardening. No pressure shall be created under the raft. This pipe network can be embedded under the floor of the basement.</li>
</ol>
<p>One more important point: place a thick layer of stones and stone dust underneath the raft. It will reduce the capillary action in the soil and water will tend to flow horizontally instead of vertically. Allow a horizontal passage of water below the raft. It can be easily done by placing half perforated pipes network below the raft in the layer of stones a d guide the water outside the building area.</p>
| 15691 | Building Uplift due to water pressure |
2017-06-05T21:07:20.047 | <p>I am not sure if this is the right place to ask this but. I have an offroad skateboard from <a href="http://mbs.com" rel="nofollow noreferrer">MBS</a>. I bought their <a href="https://www.mbs.com/parts/15005-mbs-v5-brake-system" rel="nofollow noreferrer">brake system</a> which comes with a circular piece of aluminum. Pictures attached.</p>
<p>You can see that on the circular aluminum brake part (don't know the proper name), it has 5 holes drilled. I called the company and they said they have CAD files and it is done in mass. I don't have those CAD files and I am not a CAD expert, nor do I have a machine to cut it.</p>
<p>I want to know the best way to measure (with basic relatively inexpensive tools, if needed I'll buy more), and replicate those holes with high precision on the sprocket. As you can tell, this is going to spin, so any measurement that is a little bit off is going to make the sprocket wobble. </p>
<p>The sprocket is not the same exact size as the aluminum brake part. </p>
<p>I am not looking for measurements. I am looking for how to measure. Not the fish itself, but I want to learn how to fish. </p>
<p><a href="https://i.stack.imgur.com/VgdT1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VgdT1.jpg" alt="sprocket"></a></p>
<p><a href="https://i.stack.imgur.com/ZJPQn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZJPQn.jpg" alt="wheel"></a></p>
<p><a href="https://i.stack.imgur.com/fb3zO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fb3zO.jpg" alt="inside wheel"></a></p>
| |steel|measurements|mathematics|aluminum| | <p>I am confused about why you have the gear? The center of that gear does not look like it will fit on the axle where the brake disk is. </p>
<p>If you are planning on cutting the center out of the gear and the 5 holes, I might mount it in a drill and clamp the 2 together. Spin the part with the drill in a vice or something until you an see that they are lined up correctly then you can spray paint through the holes in the brake disk to mark the gear. Depending upon the precision you need you can drill press then mill until the paint is gone. </p>
| 15693 | How to duplicate measurements from one piece of material to another? |
2017-06-06T08:41:44.173 | <p>I have a mat foundation that is 3 m thick (44 story building), I know how to design the upper layer and bottom layer but since at the top layer the reinforcement is too dense I want to make a third layer of reinforcement but my problem is I don't know how to design the third layer.</p>
<p>If for example in total in the top layer I need 55.8 cm2/m so I place T25@15 cm (32.8 cm2/m)in the top layer, for the remaining reinforcement area need I want to place them in a different layer just below the top layer bearing in mind that the effective depth will change so I can't just say 55.8-32.8 = 23 cm2/m and place them.</p>
<p>How can I calculate the remaining reinforcement needed ? </p>
| |structural-engineering|civil-engineering|design|structural-analysis| | <p>Actually, you can just subtract the area from the top layer to calculate the subsequent layers. Obviously this is an approximation, since you're losing effective depth in the other layers, but it's a good start.</p>
<p>To check that the approximation is valid, calculate the center of gravity of the total steel in each face and then calculate the requisite steel area. If it is lower than your adopted value, you're fine.</p>
<p>So, for example, if you need 55.8 cm<sup>2</sup>/m, you can adopt one layer of $\phi20/10$ (31.42 cm<sup>2</sup>/m) and another $\phi25/20$ (24.54 cm<sup>2</sup>/m), for a total of 55.96 cm<sup>2</sup>/m.</p>
<p>Assuming a concrete cover of 2 cm and a distance of 3 cm between rebar faces, your CoG moves from 3 cm (2 cm cover plus half the diameter of the $\phi20$ rebar) to </p>
<p>$$2 + \frac{\frac{2}{2}\cdot31.42 + \left(3 + \frac{2.5}{2}\right)\cdot24.54}{55.96} = 4.425\text{ cm}$$</p>
<p>You can then recalculate your mat adopting the new effective depth. If the required steel area remains below 55.96 cm<sup>2</sup>/m, you're fine. If it goes above it, just repeat the process. This works for any (reasonable) number of layers.</p>
<p>In this particular case, where you have a mat that's three meters thick, you'll be fine: the change in the effective depth (of 1.425 cm) will be little more than a rounding error.</p>
| 15711 | Mat foundation Reinforcement with 3 layers |
2017-06-06T11:37:08.053 | <p>Although I have seen a lot of literature online about how internal pressure resistance is calculated (i.e the pressure in a gas tank) I haven't found anything about the opposite scenario (i.e the pressures on a deep sea submarine).</p>
<p>How would one calculate the thickness of material needed and stresses involved at high pressures externally? Please assume a very basic level of physics knowledge (I have GCSE knowledge for sure and can probably work out and search up some of the more basic stuff, but I'm certainly no engineer).</p>
| |structural-engineering|stresses|pressure-vessel| | <p>The mathematically determined formula for a sphere with a thin wall under external pressure (which is very inaccurate - keep reading!) is simply:</p>
<p>$$ \frac{2Et^2}{r^2\sqrt{3(1-\nu^2)}} $$</p>
<p>Where $E$ is the modulus of elasticity, $t$ is the thickness, $\nu$ is the Poisson's Ratio and $r$ is the radius of the sphere (to the midplane). This is a result of Euler's method of buckling when applied to sphere's. Ignoring how we derive this formula for now, it was tested in various laboratories before use. </p>
<p>When tested, however, this formula was found to be completely inaccurate. Spheres would collapse way before this value was ever reached. It turns out that any imperfections - any flat spots, or any wrinkles would get vastly exacerbated during the collapse and cause premature failure. There was no rhyme or reason to the methods. </p>
<p>Eventually, instead of the profound mathematical $\frac{2}{\sqrt{3(1-\nu^2)}}$, the empirically determined factor for most metals was determined to be <strong>0.365</strong>. (Only <strong>32%</strong> of the maximum possible). Other materials (or even different manufacturing methods for the same material!) have different empirical values, and various regulatory agencies and scientific bodies have dedicated hundreds of people and millions of testing dollars to uncovering these factors. </p>
<p>Different shapes, with different formulas also began to creep in, with similar problems - if they were perfectly curved they could reach the theoretical values, but they would cave within about 1/3 of the maximum theoretical value!
To uncover the proper values for various shapes and materials you'd have to discuss with ASME or ASTM or other engineering organizations to find out what they've discovered. Mil Handbook 17F has a useful formula for most composites to reach close values, but it requires taking 900 computations and finding the smallest one!</p>
| 15717 | How to calculate external pressure resistance in spheres and cylinders? |
2017-06-06T13:23:31.433 | <p>We're making a vibration polisher for polishing laser-cut aluminum and steel parts, and we're looking for the abrasive medium to fill it with. We need about 30 liters of it - and the one we found (for jewelry use) was prohibitively expensive in the required amounts. We tried with several gravel kinds and the results were underwhelming.</p>
<p>What material, that would be easy to obtain and inexpensive is a common choice for a vibration polisher?</p>
| |materials|polishing| | <p>try something called sharp plaster sand. this is sold in building supply stores and is basically a high grade of beach sand with very little dirt or silt in it, and is sold in bags. it is cheap. </p>
| 15720 | Budget abrasive for vibration polishing |
2017-06-06T12:30:38.417 | <p>I am a non-engineer and I am using a gas power plant data base. Some plants listed are of the type "combustion engine", some of the type "gas turbine", some "steam turbine" and some "combined cycle". I believe to understand that combined cycle is a combination of combustion engine and steam turbine. But what is the difference between combustion engine and gas turbine then? Or does it mean the same?</p>
| |power-engineering|gas| | <p>Combustion engine - an internal combustion engine like a car motor, with pistons moved in cycles. It's inefficient but can be easily scaled down to almost arbitrarily small sizes; usually used as a backup.</p>
<p>Gas turbine - in power plants, these are very similar in construction to jet engines, where gas is the fuel - a multi-stage turbine compressor, a turbine on exhaust, high RPM; the torque produced is used to run the generator. It's not as efficient as steam or combined cycle, though more than combustion engine - but the power output can be rapidly tuned to needs, providing a response to changing demand (which is typical for the "more inert" types that use steam.)</p>
<p>Steam turbine - gas heats water in a boiler; overheated steam runs through turbines, then is cooled. This is the same principle as great most other thermal power plants (coal, nuclear, geothermal). It's usually a large installation and may take hours to get up to speed (so no rapid response to demand + waste of energy as demand rapidly vanishes) but it has a very good efficiency.</p>
<p>Combined cycle - exhaust from gas turbine (the "jet engine") is used to heat water into steam and run a steam turbine. Better efficiency than both above, and provides the much desired rapid response. Of course cost of construction is similar to sum of costs of construction of the two, and maintenance is more complex, but the operational costs are reduced.</p>
| 15721 | Difference between gas combustion engine and gas turbine? |
2017-06-06T16:40:31.173 | <p>When I watch videos of high speed trains I always see explosions of electricity near the top, or arcing. Why does that happen? I know that the <a href="https://en.wikipedia.org/wiki/Acela_Express" rel="noreferrer">Acela</a> does it a lot but other high speed trains have it, too.</p>
| |mechanical-engineering|electrical-engineering|power-engineering|rail| | <p>The upforce on a pantograph is 15-40 pounds, 60 pounds at the outside. (7-18kg, max 30 or so). </p>
<p>The trolley (contact) wire is made of solid bronze or copper, typically 4/0 to 400kcmil (107-200mm^2), with a stranded steel messenger (catenary) wire of 3/8-1/2" (10-13mm) diameter. The messenger wire is supported every 100-200 feet (30-60m) and it supports the contact wire every 6-10' (2-3m). So the contact wire is free to rise up even a foot (0.3m) as the train passes. It often has a stabilizer bar to keep it from moving laterally but is free to move vertically. </p>
<p>As discussed, any irregularity in the contact wire, or in how it's hung, can cause pantograph and wire to separate for a moment. </p>
<p>Wave action in the wire can also cause a momentary separation. Sufficient wire or train movement can cause the wire to move out onto the curved "horn" of the pantograph. </p>
<p><a href="https://i.stack.imgur.com/15Sn3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/15Sn3.jpg" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/OJ8GK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OJ8GK.jpg" alt="enter image description here"></a>
<a href="https://en.wikipedia.org/wiki/Pantograph_(transport)" rel="nofollow noreferrer">src</a></p>
<p>Irregularities in the pantograph's running surface can also cause arcing. There are typically inset copper or bronze slides; physical damage to a slide or simply a burned spot from arcing may cause the wire to lose contact. </p>
<p>Also a pantograph typically has two slides, fore and aft, and the pantograph has either linkage or strong springs to keep it level. If there is any binding or broken linkage or a fatigued or broken spring, it may not be level and may ride on its heel or toe, causing poor contact. </p>
<p>The arcing of course is caused by current. Current might remain continuous through the arc (that tendency being proportional to voltage, more likely on the high voltage systems used in high-speed rail) -- however high speed air movement is likely to snuff out the arc, severing power to the train momentarily. Talk about voltage spikes! </p>
| 15727 | Why do electric trains arc at high speeds? |
2017-06-07T17:05:38.747 | <p>I am trying to build a reed instrument played by the wind (kind of like these aeolian instruments: <a href="http://www.zmescience.com/science/physics/aeolian-harps" rel="nofollow noreferrer">Eerie musical instruments played by the wind from around the world</a> but using reeds rather than strings), but given the necessary pressure to make the reed vibrate, I am having difficulties. For example if I place a reed instrument such as a duck call near a fan, it has not enough power to drive it. I tried to place a funnel in between, but still had no luck. </p>
<p>So, I have two things to consider:</p>
<ol>
<li><p>To find the most suitable type of reed or make one from scratch (something similar to these, but all the online examples are too narrow to capture actual wind: <a href="http://www.instructables.com/id/Hypotooter-a-mini-musical-instrument" rel="nofollow noreferrer">Hypotooter -- a Mini-musical Instrument</a>) that starts vibrating in the weakest possible wind</p></li>
<li><p>To direct the wind through a funnel-like piping (maybe not a usual funnel shape, but a different shape?), so that the reed is exposed to more power, if this is a good idea at all.</p></li>
</ol>
<p>I am stuck at this point and don't even know if this is possible at all. I have heard of wind whistles or flutes played by the wind, but only when the wind very strong, however I am looking into a reed instrument because of its different sound quality + the ability for it to be played by lighter winds as well. Is this possible or is there any directions that you could suggest for me to experiment?</p>
| |fluid-mechanics|airflow|vibration|acoustics|wind-tunnels| | <p>Do you <strong>really</strong> want to make a reed instrument or a reed like instrument?</p>
<p>The reeds used in musical instruments generally are very rigid & they need a lot of air pressure on them for the reed to vibrate. A similar effect can be achieved using durable flexible materials such as a tensioned rubber like membrane/diaphragm. A single sheet of membrane can be used, or the membrane can be doubled over to form two vibrating surfaces. One of the advantages of using a membrane is that the force of air required to make a sound would be significantly less than that required for a reed.</p>
<p>A kazoo might be an initial place to start investigations.</p>
<p>Using a funnel to concentrate the air pressure is a good idea, but the size of the funnel will be important, as will be the initial strength of the wind.</p>
<p>If you do end up with something you like, you might want to consider placing it on a rotating platform, similar to a wind vane so that a change in wind direction can easily be accommodated.</p>
| 15749 | Reed played by the wind |
2017-06-08T10:30:13.990 | <p>I am using the publication <a href="https://www.diw.de/documents/publikationen/73/diw_01.c.424566.de/diw_datadoc_2013-068.pdf" rel="nofollow noreferrer">Current and Prospective Costs of Electricity Generation until 2050</a> to source ramping constraints for an economic model of three different gas power plants - CCGT, combustion turbine and steam turbine. As I'm not an engineer I have the following questions:</p>
<ol>
<li>The model is a little simplistic. For that reason, I cannot use
three different time and cost parameters for hot, warm and cold
start. For a gas power plant with load factors of between 10 and
60%, which start would you think is most relevant (i.e. most often
done): hot (shut down 8h or less), warm (8-50h) or (probably not)
cold start (more than 50h)? Would you apply something in between hot
and warm start?</li>
<li>Regarding start-up time, there are not many studies for steam
turbines (p.71 in pdf numbering). Someone advised my to use a rather
slower start-up time for steam than for combustion. So could I use a
value similar to CCGT, for example?</li>
<li>For the "ramping load gradient limit in %-Pn/min" (p.74 in pdf numbering), I have the same issue with steam turbines as described in 2).</li>
</ol>
| |gas|power-engineering| | <p>The primary advantage of gas plants is much faster ramping of output than others - they are energy-inefficient by comparison, but they are meant to handle spikes in demand in the grid. All of them will be of hot start (<8h) as even the steam ones use relatively small boilers that are quick to get up to speed, compared to conventional power plants. As result, they are kept on stand-by or operating at very low output except for when the spike needs to be handled.</p>
<p>You can definitely use the longer times found with CCGT when describing ST power plants - although <em>even longer</em> times will be found (gas burners and big boilers are cheaper to build than gas turbines; the CCGT will usually be smaller than ST plant).</p>
<p>I'd be more concerned over the inflated numbers for CCGT; they reach ~40-60% of capacity in time typical to GT plants, and then ramp up the output to 100% over the time typical for ST. That's also why you find the wildly varying values for the gradient in the document - not only is there the slow climb of steam part trailing a long way after the gas part, the power output of the gas turbine depend strongly (and non-linearly) on RPM (which again climbs at rate dependent on that power output) - generally, first 30% or so will be rather slow, then the engine picks up and the 50%-80% ramp is really steep. Regardless, 5-10 minutes is an adequate estimate of the time from ignition to full power (not sure what preparations the operator must perform beforehand). And then water in the steam part only starts heating slowly, steam turbine a long way from budging...</p>
<p>Any kinds of averages or momentary measurements in CCGT are seriously misguiding due to that awfully non-linear output ramp.</p>
<p>Similarly, for GT, the 100% output time and climb rate will depend strongly on each other. If you allow the GT to spin up with coils off, all its power directed towards increasing RPM, it will reach peak torque RPM quickly, and provide 100% output in a really short time since. OTOH if you ramp it up while drawing power, the extra load will reduce climb rate of RPM and delay reaching 100% output. So, a choice between 5 minutes of nothing, then 100%, or 12 minutes of output that is maybe 20% over the first 7 minutes, and then climbs to 100% over the remainder of the time.</p>
| 15760 | What ramping constraints for gas plant - CCGT, combustion and steam turbine? |
2017-06-08T19:01:03.437 | <p>I ran across the diagram below in the <a href="https://i.stack.imgur.com/fvlRy.png" rel="nofollow noreferrer">mindtomachine blog</a> post describing an homebrew arc welder. It shows two transformers with inputs in parallel and outputs in series. Is there some advantage to this arrangement rather than a single transformer with a doubled turns ratio?</p>
<p>Perhaps it's just what he had at hand, but I wanted to confirm with someone having EE background. </p>
<p><a href="https://i.stack.imgur.com/5a829.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5a829.png" alt="Welder Circuit"></a></p>
<p>P.S. I'm <em>not</em> planning on building this, just curious.</p>
| |electrical-engineering|power-electronics| | <p>Doubling the turns ratio is not exactly the same.</p>
<p>Yes, you get double the output voltage but you get different Magnetizing Inductance. </p>
<p>Moreover, the real reason to do something like that could probably be related to heat dissipation. The more wire layers you put, the hotter the center of the core will be for a given power. You could run thicker wire to offset this, but then you run out of core to wind. This could be specifically true for an arc welder in which voltage is actually reduced and you run very very high currents on the output.</p>
| 15769 | Transformers in parallel/series |
2017-06-09T02:33:01.530 | <p>2-stroke engines seem to offer a massive advantage over 4-stroke engines. In the case of gasoline engines, there is a need for forced induction, but in the case of large diesels (which invariably have forced induction anyway) it seems that their advantages outweigh the disadvantages.</p>
<p>So why is it that all newly developed diesels (not counting large ship engines and research designs such as Achates Power) are 4-stroke?</p>
| |automotive-engineering|diesel| | <h2>It's all about what comes out</h2>
<p>Two-stroke, forced-induction diesels have been a mainstay of heavy industry for quite some time (Detroit Diesel 71 series, EMD 567/645/715, never mind ship engines). However, they aren't as amenable to precise emissions control as their four-stroke counterparts. This can be shown by EMD's switch to the four-stroke J (EMD 1010) engine for the SD70ACe-T4 locomotive, as they were unable to get the EMD 715 to meet EPA Tier 4 without eating an unacceptable weight penalty or using SCR (which is a logistical problem for railroads, having to deal with large quantities of yet <em>another</em> consumable).</p>
| 15775 | Why are 4-stroke diesel engines preferred in new developments? |
2017-06-09T20:54:14.887 | <p>In order to find the coordinates of the center of an area, we can use this formula:
$$\begin{align}
\overline{x} &= \frac{\int x\text{d}A}{A} \\
\overline{y} &= \frac{\int y\text{d}A}{A}
\end{align}$$</p>
<p>How can we derive this formula? Can you help me understand it so it becomes intuitive?</p>
| |statics| | <p>Let me suggest a different approach:
Let <span class="math-container">$\Omega$</span> be an object in 2D.
Let <span class="math-container">${\bf x} = (x,y)$</span> be a point in the plane. Let <span class="math-container">${\bf \bar{x}}=(\bar{x},\bar{y})$</span> be the center of mass of <span class="math-container">$\Omega$</span> and let <span class="math-container">$\rho(x,y)$</span> be mass density function (mass per unit area). The force density <span class="math-container">${\bf F}$</span> can be obtained by multiplying <span class="math-container">$\rho$</span> by the gravitational acceleration <span class="math-container">${\bf g}$</span>, which points downwards. Then, since <span class="math-container">${\bf \bar{x}}$</span> is the center of mass, by static equilibrium, the sum of the contributions of all the torques with respect to <span class="math-container">${\bf \bar{x}}$</span> is zero. Mathematically,
<span class="math-container">$$
\begin{align}
{\bf 0} &=
\int_{\Omega} {\boldsymbol{\tau}} \,d\Omega \\
&= \int_{\Omega} \overbrace{({\bf {x}}-{\bf \bar{x}})}^{\bf r}\times
\overbrace
{
{\bf g} \, \rho(x,y)
}^{\bf F} \, d\Omega \\
& =
\int_{\Omega} ({\bf {x}}-{\bf \bar{x}})^{\perp}\,g \,\rho(x,y)\, d\Omega,
\end{align}
$$</span>
where <span class="math-container">$({\bf {x}}-{\bf \bar{x}})^{\perp}= ((y-\bar{y}),-(x-\bar{x}),0)$</span> is the vector <span class="math-container">$({\bf {x}}-{\bf \bar{x}})$</span> rotated <span class="math-container">$\pi/2$</span> radians CCW.
This is a consequence of the cross product and the fact that the vector <span class="math-container">$({\bf {x}}-{\bf \bar{x}})$</span> is perpendicular to <span class="math-container">${\bf g}$</span>. Hence, equating vector component yields
<span class="math-container">$$
\begin{align}
0 &= \int_{\Omega} {(y-\bar{y})} \,\rho(x,y)\, d\Omega \\
0 &= \int_{\Omega} {(x-\bar{x})} \,\rho(x,y)\, d\Omega
\end{align}
$$</span>
which implies that
<span class="math-container">$$
{\bar{x}} = \dfrac
{
\int_{\Omega} { {x}} \,\rho(x,y)\, d\Omega
}
{
\int_{\Omega} \,\rho(x,y)\, d\Omega
}
$$</span>
<span class="math-container">$$
{\bar{y}} = \dfrac
{
\int_{\Omega} { {y}} \,\rho(x,y)\, d\Omega
}
{
\int_{\Omega} \,\rho(x,y)\, d\Omega
}
$$</span>
Lastly, if your density is constant, we obtain the original result.</p>
| 15786 | How can we derive the formula of the centroid? |
2017-06-10T09:28:31.310 | <p>As you know in linear theory of elasticity equations for strains are given as (according to picture):
<a href="https://i.stack.imgur.com/4pGfL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4pGfL.png" alt="enter image description here"></a></p>
<p>But what will be if an element will just rotate (strain equation is on the picture):
<a href="https://i.stack.imgur.com/0mBo7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0mBo7.png" alt="rotation of an element"></a></p>
<p>Intuitively i think there should be no strains. The main question is how to deal with it? </p>
| |mechanical-engineering|structural-engineering|structural-analysis|stresses|elastic-modulus| | <p>From your picture, your deformation map would be something like</p>
<p>$$
\varphi(x) =
\left[\begin{array}{c}
xcos(\varphi) - ysin(\varphi) \\
xsin(\varphi) + ycos(\varphi) \\
z \\
\end{array}\right]
$$
Deformation gradient
$$
F =
\left[\begin{array}{ccc}
cos(\varphi) & -sin(\varphi) & 0 \\
sin(\varphi) & cos(\varphi) & 0 \\
0 & 0 & 1\\
\end{array}\right]
$$
Green deformation tensor
$$
G =
\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1\\
\end{array}\right]
$$
$$
\therefore
E = \frac{1}{2} [G - I] = 0
$$
Which is what we would expect, but in order to calculate this we needed to use higher order terms. Since:
$$
G = FF^T = (\nabla u + I)(\nabla u + I)^T = \nabla u + \nabla u^T + \nabla u \nabla u ^T + I
$$
If each component of:
$$
\nabla u < 1
$$
then each component of:
$$
\nabla u \nabla u ^T \ll 1
$$
We expect our $\varphi$ to be very small, so we can ignore our higher order terms. Let's linearize the deformation gradient:
$$
cos(\varphi) \approx 1, sin(\varphi) \approx \varphi \\
F =
\left[\begin{array}{ccc}
1 & -\varphi & 0 \\
\varphi & 1 & 0 \\
0 & 0 & 1\\
\end{array}\right] \\
G =
\left[\begin{array}{ccc}
1 + \varphi ^2 & 0 & 0 \\
0 & 1 + \varphi ^2 & 0 \\
0 & 0 & 1\\
\end{array}\right] \\
\therefore
E = \frac{1}{2}
\left[\begin{array}{ccc}
\varphi ^2 & 0 & 0 \\
0 & \varphi ^2 & 0 \\
0 & 0 & 0\\
\end{array}\right] \approx 0
$$
This makes since, because if $\varphi$ is small, $\varphi^2$ is much smaller and we can accept the error assuming geometrically linear behavior. We just need to watch out for non-small rotations that will cause our linearization to have a larger error.</p>
<hr>
<p>Resource:</p>
<p>Fundamentals of Structural Mechanics 2nd, Keith D. Hjelmstad</p>
| 15788 | How to deal with rotation of an element in theory of elasticity? |
2017-06-10T21:20:48.730 | <p>I tried to deliver a refrigerator and the customer saw that the refrigerator was lying down in the truck instead of standing up. He refused to accept it, claiming that the compressor would be damaged by having it on its side. I tried to explain that that made no sense, but his friend came to me and also said the same thing. </p>
<p>Is transporting a refrigerator on its side a problem? If so, why?</p>
| |refrigeration| | <p>You can lie the fridge On the side with all the connection pipes of the compressor pointing upwards. (Ignoring the crimped off process pipe.)<br />
If there are pipes on both sides, lie it with the bigger diameter one pointing up.
It doesn't actually matter too much. The object is to stop compressor oil getting into the circuit and then needing to be blown all the way around by the refrigerant. If that happens, it should come right by itself but can take a while (hours)</p>
<p>I got this advice from a seinor production engineer at a New Zealand refrigeration manufacturing plant (Dad). <em>If</em> fridges are not made the same way everywhere YMMV.</p>
<p>I'll leave it to your common sense how to secure the door for handling and transport.</p>
| 15798 | What's wrong with transporting a refrigerator on its side? |
2017-06-11T08:35:17.263 | <p>I am trying to understand / interpret one recommendation I read on a Caterpillar website for maximizing excavator life:</p>
<p>**</p>
<blockquote>
<p>Never dig over final drive.</p>
</blockquote>
<p>**</p>
<p>Can anyone explain what this means? What's the "final drive" and what does digging over it mean?</p>
<p><a href="https://i.stack.imgur.com/NSYoS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NSYoS.jpg" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/zKWqN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zKWqN.png" alt="enter image description here"></a></p>
| |mechanical-engineering|equipment-selection| | <p>The final drive is the planetary gear set and hydraulic motor that move the track. Not digging over it means not operating the arm over that area as it stresses the components unnecessarily - so to avoid this it means positioning the machine correctly before starting work or changing its position during work.</p>
<p><a href="https://i.stack.imgur.com/xyjQh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xyjQh.jpg" alt="final drive"></a></p>
| 15803 | Understanding a Caterpillar Excavator Recomendation for Increasing Life |
2017-06-11T17:23:30.977 | <p>Enthalpy seems to be commonly used to characterize geothermal reservoirs, but how is it calculated? I have an example here where the enthalpy is cited as 800 $kJ kg^{-1}$ for a reservoir with a temperature of 190 °C. In that case the enthalpy was apparently just taken to be the specific heat of water (4200 $J kg^{-1} °C^{-1}$) multiplied with the temperature, something like $H = c\Delta T$ (meaning that the reference temperature must be 0 °C).</p>
<p>Is that the correct way to do it? And if yes, is this really the enthalpy in the thermodynamical sense and why is the reference temperature 0 °C? </p>
| |thermodynamics|energy|renewable-energy| | <p><em>Nitpicky</em> Answer: <strong>you can't</strong>.</p>
<p><em>More Detailed Answer</em>: While <a href="https://en.wikipedia.org/wiki/Enthalpy" rel="nofollow noreferrer">enthalpy</a> is commonly treated as an absolute quantity it is a <a href="https://en.wikipedia.org/wiki/Thermodynamic_potential" rel="nofollow noreferrer">thermodynamic potential</a>. This is why textbooks might show the definition as <a href="https://itp.uni-frankfurt.de/~valenti/WS13-14/all_1314_chap5.pdf" rel="nofollow noreferrer">\1\</a>:</p>
<p>${\displaystyle \mathrm{d}H=\mathrm{d}U+\mathrm{d}\left(pV\right)}$</p>
<p>This means (essentially) you need a reference-point and all values for "enthalpy" are the differences in <em>thermodynamic potential</em> with respect to this reference point. </p>
<p>Let's illustrate this with a thought experiment. Imagine a reservoir (artificial lake behind a dam) and let's pretend it contains 1000l. The question how much energy can be produced by a pelton-turbine can only be answered when you add information about the height (In this case its even called potential energy).</p>
<p>In order to quantify the thermodynamic potential of a geothermal resource it is necessary to define a meaningful reference point and this reference point needs to be disclosed otherwise the numbers are not meaningful.</p>
<p>\1\ <a href="https://itp.uni-frankfurt.de/~valenti/" rel="nofollow noreferrer">https://itp.uni-frankfurt.de/~valenti/</a></p>
| 15814 | How do you calculate the Enthalpy of a geothermal resource? |
2017-06-12T11:03:06.570 | <p>I want to cast items out of epoxy resin. They will be tools used to handle ESD sensitive devices, so they need to at dissipate static charge, or perhaps somewhat conduct. Resins used will vary depending on the part.</p>
<p>Unfortunately, resin suppliers seem to never give any data on electrical conductivity. I presume it's extremely low. So to have it conduct, am I right in saying I could mix in something like graphite to help conduction? I presume it would reduce strength, or at least change some physical properties, but that's fine.</p>
<p>I've previously experimented mixing graphite powder with non-conductive powders - No liquids or glue, and just compacted by hand in a glass tube. I found that the resistivity could not be controlled easily / effectively by varying the ratio; it seemed it either conducted quite well, or not at all, to the degree where fine tuning ratios would too unreliable. </p>
<p>Might that be the case with epoxy resin as well? How might I increase the conductivity of a resin? And how might it be tunable? Principles / formulas? I presume the geometry, size and other physical properties of the powder used is important, as well as, perhaps, the resin?</p>
| |composite-resin| | <p>For non conductive samples in an electron microscope , silver paint was used to conduct away any charge. I would expect copper powder or small fibers may be available also. Caution , I tried mixing fine zinc powder into an epoxy and it immediately reacted with the liquid epoxy. </p>
| 15818 | How to make epoxy resins dissipate static charge |
2017-06-13T05:29:15.587 | <p>I wonder how can I make hard disk drive more durable? Specifically, I works in a factory environment. And almost every few months, hard disk drive in some of the factory machine are corrupted, and even unrepairable. We already implement several SSD on some of them, but that's not much of help. They're being corrupted too on factory machine with heavy task.</p>
<p>So every time it happened, we always restoring using our backup image. And we already attach all of the hard disk drive with shock absorber to reduce the shake.</p>
<p>Is there any other option or prevention we could use? Perhaps adding any anti magnet material to prevent magnet friction, or something else? And what material we need?</p>
<p><strong>Edit:</strong>
Factory machine with heavy task I mention above basically machine to create car metal body and car frame mould.</p>
<p>And when I meant the disk is corrupted, it means unreadable. The whole disk. Not the program files or software related. So it won't boot at all.</p>
| |mechanical-engineering|machining|computer-hardware| | <p>Really a comment but too long:</p>
<p>I've dealt with PCs on the factory floor (woodworking), they proved quite resilient.</p>
<p>Our initial setup which was basically trouble free: We mounted the PCs inside a cabinet, the front was clear plastic slats (think what you sometimes see on a walk-in refrigerator freezer). The original intent was to maintain a slight positive pressure with clean air but this was never done and proved to not be needed.</p>
<p>Unfortunately, after that plenty of machines got installed with less care. The usual "failure" mode was thermal shutdown, take the cover off and blow it out, it would work fine, although these did prove more problematic as the dust did some damage.</p>
<p>The main problems, however, came from their wires. We specced shielded cables but management went cheap on us, the building was wired with ordinary network wire and later modifications were often made by electricians rather than computer guys. This caused a <strong>lot</strong> of interference and was probably responsible for the high failure rate of the network cards. (Really, now, a Cat-5 tossed over a 480V, 400?A main power bus??? Or even more extreme, a Y connection in a cat 5--which actually worked, albeit with network error problems!) Don't put a computer on the same circuit as a heavy motor. Don't run any computer wires parallel to heavy power wires even if they're on separate circuits.</p>
<p>The initial machines were all diskless (not an option these days) and even after that everything of importance was stored on the network so if a machine did act up it could be swapped out very quickly--it took longer to carry the new machine to the station than to get it up and running in place of the problematic one.</p>
<p>Top lesson--don't let the electricians be anything but carefully-watched assistants when wiring things.</p>
| 15832 | How to make factory machine hard disk drive more durable? How much its estimate life span? |
2017-06-13T08:14:56.023 | <p>I have read that variable speed hydraulic drives have very low part-load efficiency - due to losses in the internal pumps and motors.</p>
<p>What is part-load?</p>
| |mechanical-engineering|fluid-mechanics| | <p>Part-load is when the full power possible of the drive / motor is not used or needed. The efficiency is lower due to the losses becoming more significant compared to the lower power produced as some of the losses may be independent of power produced.</p>
| 15835 | Part load definition with respect to hydraulics |
2017-06-13T14:20:37.233 | <p>I was checking the <a href="https://en.wikipedia.org/wiki/Mobile_country_code#India_-_IN" rel="nofollow noreferrer">list of MCC (Mobile Country Codes) on Wikipedia</a> and found that India has two MCC: 404 and 405. 404 contains only GSM/UMTS/LTE operator bands and 405 contains both GSM/UMTS/LTE and CDMA operator bands. The problem is that I do not understand the reason for having 2 MCC for the same country. We also have mobile network code (MNC) to differentiate between various operators, circles and the bands.</p>
| |telecommunication|wireless-communication| | <p>The Mobile Network Code (MNC) is either a two digit ("European" standard) or three digit ("North American" standard) identifier, with in the Mobile Country Code (MCC).</p>
<p>According to <a href="https://en.wikipedia.org/wiki/International_mobile_subscriber_identity" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/International_mobile_subscriber_identity</a>, the 3GPP specification 23.003 recommends that the "length of the MNC is uniform within a given MCC area".</p>
<p>Looking at the OP's link, presumably India was allocated MCC = 404 and started using two-digit MNC numbers, but then came close to running out of numbers. The additional MCC = 405 has blocks of three-digit numbers as well as two-digit, so they may be unlikely to ever need another MCC allocation.</p>
| 15846 | Why does India have more than 1 MCC (Mobile Country Code)? |
2017-06-16T06:49:23.337 | <p>I bought sunglasses in China that claims to have UV400 protection. I'd like to test this as my eyes' health is on the line :)</p>
<p>Is there any way to check the maximum UV amount these glasses can protect me against?</p>
| |product-testing| | <p>Someone proposed an original method. The result is binary (it doesnt give you a percentage) but it might gives you a rough idea if UV pass through the glass or not: </p>
<blockquote>
<p>Testing can be one with a simple trick that will require access to a
UV flashlight. You can find an affordable UV flashlight online or
borrow one. Once you have it, go with it into a darkroom together with
the sunglasses and paper money or a credit card. Turn on the
flashlight and beam it on the credit card or money. Various strange
symbols that you do not usually see in ordinary light will become
visible. They include a line on the paper money and letters on credit
cards. These are watermarks that the government and banks use to
identify counterfeits.</p>
</blockquote>
<p>source <a href="https://www.quora.com/How-can-I-tell-if-my-sunglasses-are-UV-protected/answer/Ronak-Furia-2" rel="nofollow noreferrer">Ronak Furia, on quora</a></p>
| 15884 | Test UV protection on sunglasses |
2017-06-17T09:02:58.393 | <p>when designing a pond it is customary to produce a chart for the local council, specifying the height of the water for the volume stored. What is the name of this chart?</p>
| |civil-engineering|water-resources| | <p>Storage curve or water supply curve are both possible depending on the planned use.</p>
| 15907 | Chart that describes how full the pond is |
2017-06-17T15:48:08.220 | <p>I need to convert 7.62cm/3in linear motion from an actuator to 150-180° (150 is minimum) rotary motion. I am aware of the existence of gears, but would probe here to see if there is some simpler method. I could use rods mounted in some way. It has to be compact and rugged, however. I'd also be interested in formulas for this type of conversion. Thanks! </p>
<p>Edit:
I found this contraption <a href="https://i.stack.imgur.com/le6em.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/le6em.gif" alt="Mystery device"></a> for microlinear precision, whilst googling for scotch yoke. Is this a scotch yoke? If not, what is it? I can probably awfulhack one of these with a file, some aluminium rods and strips. But it would be prettier if I bought one.</p>
| |mechanical-engineering|actuator| | <p>As suggested a rack and pinion will provide a linear transfer function and avoids complex linkages. A variant on the rack and pinion is to replace the straight rack teeth with a taught cable and the pinion teeth with a pulley. This mechanism is sometimes referred to as a capstan and bowstring.</p>
<p><a href="https://en.wikipedia.org/wiki/Capstan_and_Bowstring" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Capstan_and_Bowstring</a></p>
<p>If you need a guarantee against slippage then the cable can be wrapped multiple times and attached to the pulley at the half-way point.</p>
| 15909 | How to convert 7.62cm linear motion to 150-180° rotary motion |
2017-06-17T17:39:25.960 | <p>I have this skateboard truck for my off-road skateboard. I want to mig weld a piece of aluminum to it to mount a motor. How would I test if it is something I can weld to with a MIG welder? I think it is aluminum, how can I be sure?</p>
<p>From what I understand</p>
<ul>
<li><p>if magnets don't stick to it, it is aluminum. </p></li>
<li><p>if I test for continuity with my multimeter and I don't have continuity, it is probably anodized. </p></li>
<li><p>if it were polished it would be very shiny and probably reflect light. </p></li>
</ul>
<p>How do tell if it is in fact annodized, and if so, what do I need in order to weld through it? Do I sand it down? I was going to use a flux core mig welder.</p>
<p><img src="https://i.stack.imgur.com/17MF1.jpg" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/kOh3G.jpg" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/xfwaQ.jpg" alt="enter image description here"></p>
<p>For reflection</p>
<p>Also here is the inside where they have an existing weld.
<a href="https://i.stack.imgur.com/NtGsC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NtGsC.jpg" alt="enter image description here"></a></p>
| |metallurgy|aluminum|welding| | <p>That does look a lot like aluminium. All grades of aluminium are non-magnetic as are some stainless steels. The best test is weight as aluminium is significantly less dense than steel but has lower tensile strength so judging by the bulk aluminium seems likely. a final test is that if you grind aluminium with an abrasive wheel it will produce virtually no sparks. </p>
<p>I've not heard of flux core wire for aluminium, stick rods for aluminium alloys do exist but tend not to give great results in general. </p>
<p>Generally TIG is the best option for aluminium alloy welding but a decent MIG machine with pure argon shielding gas and the appropriate filler wire can produce good results. The difficulty with MIG is that aluminium wire is significantly softer than steel so tends to have more problems with wire feed. </p>
<p>Another issue is that not all grades of aluminium have good weldability t<a href="http://www.lincolnelectric.com/en-gb/support/welding-solutions/Pages/aluminum-faqs-detail.aspx" rel="nofollow noreferrer">his article</a> gives good overview. </p>
<p>The other issue, which applies to welding any finished part is that the welding process inevitably puts a significant amount of heat into it which can affect its mechanical properties on a material level an also cause dimensional distortion, which can be mitigated but never entirely eliminated. </p>
<p>The short answer is that it may not be impossible to do but is likely to be considerably more difficult than just welding a bracket to bit of mild steel and modifying finished parts by welding is always a bit risky when you don't know the specific of their material and manufacturing process. </p>
| 15911 | Can I mig weld to my skateboard truck? How can I test if it's weldable? |
2017-06-20T13:13:04.043 | <p>As a complete novice, I have been reading about outside engineering projects such as decking and patios. </p>
<p>All the information I read tells me that to lay structural joists as a subframe / sleeper joists, the joist should be positioned with the thin side down, the wider side running perpendicular to the ground. </p>
<p>From my understanding, this makes for a stronger frame which can take more weight. My question is does it <em>have</em> to be this way? For example, if the goal was to create a lower deck, is it structurally sound for the joists to be laid "wide side" down?</p>
| |structural-engineering| | <p>Beams have a property we call moment of inertia ($I$) which controls how strong they are to resist bending. For rectangular beams, the equation is</p>
<p>$$I = \frac{bh^3}{12}$$</p>
<p>where $b$ is the cross-section's base and $h$, its height. This means that the height of the cross-section is far more important than the base. Indeed, if you need a 2x6" joist "thin-side-down", then you'd need a 2x54" (yeah, that's not a typo) joist "wide-side-down" to withstand the same load. The two sections below have the same strength against bending:</p>
<p><a href="https://i.stack.imgur.com/zc8Za.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zc8Za.png" alt="enter image description here"></a></p>
<p>And obviously, this doesn't even take into consideration that joists are usually spaced some 16" apart, which is impossible with the necessary wide-side-down section (since it's 54" wide).</p>
| 15942 | Does a joist have to lay "thin side" down? |
2017-06-20T13:35:29.547 | <p>I am about to present my work on vibrations to a NVH specialist at work and he is supposed to help me.</p>
<p>I have calculated PSD's, now my question (subjective, I know) is how should I plot the data? I've been plotting $PSD \big[\frac{g^2}{Hz}\big]$ against frequency $\big[Hz\big]$, both in linear axes. I've been told the frequency axis should be in logarithmic scale, which makes sense, what about the PSD axis?</p>
| |mechanical-engineering|vibration| | <p>Typically, a PSD for vibration will have a log scale for both axes. </p>
| 15944 | Best format to plot vibration data? |
2017-06-20T18:58:10.213 | <p>Let's start with arbitrary transfer function as an example, such as:</p>
<pre><code>obj = tf([1 2],[1 0 2 5]);
nyquist(obj);
grid on;
axis([-2 2 -2 2]); axis equal
</code></pre>
<p>As far as I do understand what is gain (expressed in dB), I am unable to figure out, nor find any information about circles that appear after we use <code>grid</code>. For some reason there is vertical line at <code>Re=-0.5</code> associated with <code>0 dB</code> value. On the left and right of this line we see non coincident circles in the shape of peacock's eye with decreasing or increasing values. Circles converge to points <code>[-1,0]</code> and <code>[0,0]</code>. </p>
<p>We can see that system's curve starts at point <code>[-0.4,0]</code> which corresponds to <code>-7.9588 dB</code> gain, crosses <code>-10 dB</code> circle twice, where on both occurrences system's gain doesn't match <code>-10 dB</code> value.</p>
<p>Moreover, just from the fact that that we have infinite line denoted with <code>0 dB</code> value, one can infer that it has absolutely nothing to do with actual gain, which on Nyquist plot should be circular around point <code>[0,0]</code>. In this case my question is simple: what does those dB values and circles mean?</p>
<p><a href="https://i.stack.imgur.com/5SWqB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5SWqB.png" alt="Nyquist plot with grid on"></a></p>
| |control-theory|matlab| | <p>See <a href="https://www.mathworks.com/help/control/ref/nyquist.html" rel="noreferrer"><code>doc nyquist</code></a>:</p>
<blockquote>
<p>The nyquist function has support for M-circles, which are the contours of
the constant closed-loop magnitude. M-circles are defined as the locus of
complex numbers where</p>
<p>$$T(j\omega) = \left|\dfrac{G(j\omega)}{(1+G(j\omega))}\right|$$</p>
<p>is a constant value. In this equation, $\omega$ is the frequency in
radians/TimeUnit, where TimeUnit is the system time units, and $G$ is the
collection of complex numbers that satisfy the constant magnitude
requirement.</p>
</blockquote>
| 15952 | Nyquist plot - what is the meaning of circles with dB values on complex plane |
2017-06-20T19:49:48.417 | <p>Our house air conditioner went out last Saturday afternoon on a +100 F day. Was able to diagnose the problem itself as a problem with the starting circuit of the big compressor fan of the outside compressor unit. If I gave the fan a strong push with a stick to start it rotating, then the fan motor would start up and the air conditioning would work fine. An AC technician later determined that the compressor fan needed a new start-up capacitor, and now the AC works fine. </p>
<p>My question is this: Why are large 2-ton house air conditioners designed so that their compressor fans require start-up capacitors to turn on? The diameter of the compressor fan on our house AC appears to be roughly the same as that of an inexpensive box fan that I can get at Walmart, and the ac motors on those simple box fans don't require a start-up capacitor to turn on, do they? You just plug them in and they smoothly start revving up to speed all by themselves. So why do house air conditioners use compressor fans that need these trouble prone start-up capacitors, and why can't a simple self-starting ac motor like that of a box fan be used instead for the compressor fan?</p>
| |electrical-engineering|motors| | <p>An <a href="https://en.wikipedia.org/wiki/Induction_motor" rel="nofollow noreferrer">induction motor</a> works by trying to sync up the rotating squirrel cage to the sine wave of the mains voltage.</p>
<p>With a single phase supply there is no difference between going forwards or backwards so you need some way of setting the direction on startup. </p>
<p>For low torque applications a <a href="https://en.wikipedia.org/wiki/Shaded-pole_motor" rel="nofollow noreferrer">shaded pole</a> is enough. That is a small loop of copper on the core placed in such a way that the magnetic field gives a push in the right direction. </p>
<p>For higher torques they instead have a secondary winding perpendicular to the main winding. A capacitor is used to create a <a href="https://en.wikipedia.org/wiki/LC_circuit" rel="nofollow noreferrer">LC circuit</a> to run the secondary winding slightly out of phase with the main winding. Once the motor is up to speed a centrifugal switch may be used to disconnect the secondary winding and starter capacitor which are often not built for continuous use.</p>
| 15954 | Why does the Compressor Fan for an Air Conditioner need a Starter Capacitor? |
2017-06-21T00:41:23.523 | <p>Concrete/brick homes in tropical areas are hot due to direct sunlight. Please suggest external siding materials or design to give heat shielding. Please give reasons (and if possible pros & cons).
Some ideas are using wood siding, because wood thermal capacity is low so that any heat will blown away by wind (but interstitial space behind the wood may infested with animals). Or using living plant leaves, since they give shade and heat on leaves blown away by wind (but may have problems with mould growth, or plant maintenance).</p>
| |materials|heat-transfer|architecture| | <p>Use an over-hanging roof which will then shade the walls. Over-hanging roofs are seen in many countries, but not only for sun but also for snow (eg Switzerland) to stop or reduce the chance of snow piling up against the wooden chalet walls </p>
<p>This type of design is called "passive solar design" and there are many links / books that can be found.</p>
| 15957 | House / home external skin for heat shield materials |
2017-06-21T22:15:13.990 | <p>How does the trigger of a typical modern multi speed drill (maybe Ryobi for example) work? Is it just a potentiometer? Or is it more involved?</p>
| |electrical-engineering| | <p>I opened a Ryobi type of cordless drill. The control appeared to be a potentiometer connected to a PWM circuit module which varied the duty cycle of the power applied to the motor.</p>
| 15973 | Is the trigger of a drill a potentiometer? |
2017-06-22T06:30:45.217 | <p>Recently I visited a sericulture facility in India and I saw that people there 'degummed' (<a href="https://www.youtube.com/watch?v=h13fctiTylI" rel="nofollow noreferrer">https://www.youtube.com/watch?v=h13fctiTylI</a>) silk in a batch process. </p>
<p>As a Chemical Engineer, I thought of this as an unsteady state unit operation and
I have decided to work on making this process a continuous or steady-state operation as a summer project.</p>
<p>So far this is my schematic
<a href="https://i.stack.imgur.com/ndkJM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ndkJM.jpg" alt="enter image description here"></a></p>
<p>The silk thread will be pulled over a large drum and will pass over this drum a number of times in a helical fashion. This drum is immersed in a bath of boiling water with added chemicals so that the silk is degummed. The thread will come out of the other end another end of the cylinder and will be wound on another bobbin. </p>
<p>Now I have already determined what sort of chemicals and mass transfer I am gonna get and what should be my temperature of the bath. I am not able to work out on the residence time any part of the silk thread should spend in the bath ? for this I need to determine the RPM of the rotating drum? how do I calculate that? standard degumming time is about 60 minutes. So if I say that the length of the drum is L and its diameter is D and say it is rotating at an angular velocity of w. How do I develop a set of equations to determine my rpm so that I can change my drum dimension if the residence time is too long or too short.</p>
| |mechanical-engineering|chemical-engineering|applied-mechanics| | <p>Length $l$ of silk on the immersed part of the drum $m$ divided by feed rate $f$ is the time the thread is immersed <em>T</em>. </p>
<p>$l / f = T$. </p>
<p>Feed rate is same as linear velocity of the surface of the drum (assuming negligible silk thickness comparing to drum radius). $f = v = \omega r$. </p>
<p>The amount of thread on the drum ($l'$) is the drum circumference ($C$) times number of loops ($n$); number of loops is the usable length of the drum ($L$) divided by pitch of the helix ($p$; which must be no smaller than the silk thickness, preferably at least 2x as large). </p>
<p>$l' = nC$; $n = L/p$ ; $C = 2 \pi r$ ; $l' = 2\pi r L/p$</p>
<p>If the whole drum is immersed, $ l = l'$ so, </p>
<p>$$ {{2\pi r L/p} \over {\omega r}} = T$$</p>
<p>or,
$$ \omega = {{ 2 \pi L } \over {pT}} $$ </p>
<p>where the drum of length L and helix pitch p rotating at $\omega$ holds the thread immersed for time T. </p>
<p>(Interestingly, makes the radius non-issue). </p>
<p>If your drum is partially immersed, you'll need to use the <a href="https://en.wikipedia.org/wiki/Circular_segment" rel="nofollow noreferrer">Circular Segment</a> length to determine percentage of the circumference that remains submerged, and adjust the time accordingly.</p>
| 15976 | I need to determine the RPM of a rotating cylinder over which a silk thread is wound and is moving along in the fashion of a screw. |
2017-06-22T12:33:27.103 | <p><a href="https://www.youtube.com/watch?v=RIXSzw2ksiE" rel="noreferrer">Soap bubble foam made with helium</a> floats up, but due to extreme fragility hardly counts as "material". There are many solid foam materials though - PUR foam, or styrofoam to name the most common. They typically use carbon dioxide for inflation though (usually produced from precursors of the foam, as a desirable side effect of their reaction). </p>
<p>But it shouldn't be too difficult to make solid foam filled with helium (or hydrogen) in proportions assuring positive buoyancy in air, and I can imagine desirability of it, at least as a filler in applications where mass costs a premium (transport, aviation) even if its structural properties were to be too poor for any other purpose.</p>
<p>Is such material produced? Is it used anywhere? Or if not, why?</p>
| |materials| | <p>Here is a link to an aerogel <a href="https://www.extremetech.com/extreme/153063-graphene-aerogel-is-seven-times-lighter-than-air-can-balance-on-a-blade-of-grass" rel="nofollow noreferrer">7 times lighter than air</a>.</p>
<p>They are mainly in experimental phase. To make them into practice, they should be cheaper as the equivalent system filled with Helium.</p>
<p>They are mainly porous materials derived from a <a href="https://en.wikipedia.org/wiki/Aerogel" rel="nofollow noreferrer">gel</a>, from which the liquid component has been removed.</p>
<p>The result is like this (it is a different aerogel as in the first link):</p>
<p><a href="https://i.stack.imgur.com/uE7kcm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uE7kcm.png" alt="enter image description here"></a></p>
| 15977 | Is there a lighter-than-air foam material? |
2017-06-22T21:30:26.747 | <p>When creating a detention pond I have only came across examples where the pond is created by digging. Is it possible that a pond is created by creating retaining walls / embankments? and if it is can you please provide examples.</p>
<p>Please note! this is a detention pond not decorative pond.</p>
| |civil-engineering|water-resources| | <p>In my experience you should always dig when creating detention ponds. This allows you to position it at the best location available to you incase water levels run high. </p>
<p>For this purpose, you may want to build the banks up higher than you actually intend them to go, and then leave a gap where you plan for any excess water to escape. You'll need to create a path leading away from the pond toward a creek, stream or river. You'll also need the local government's permission to do so before you introduce a new tributary to the local water table.</p>
<p>Depending on how big the pond is I would also recommend a very good PVC liner or if ita a bigger detention pond some kind of <a href="https://tankliners.com.au/" rel="nofollow noreferrer">tank liners</a> will do the trick.</p>
<p>I found a really good guide here also: <a href="https://www.hunker.com/13424245/how-to-create-a-retention-pond-for-water-runoff" rel="nofollow noreferrer">https://www.hunker.com/13424245/how-to-create-a-retention-pond-for-water-runoff</a></p>
<p>good luck with your project</p>
| 15991 | Are detention ponds always dug? |
2017-06-25T03:12:18.473 | <p>I am wondering if there is a formula (or how to derive one) for calculating the minimum radius a rod (with a given radius/diameter) can be bent around while staying in the elastic range. Obviously, this will depend on the modulus of elasticity for the material.</p>
<p>I am searching for a general formula for this rather than a specific solution for a particular material. With a general formula, I can evaluate trade offs using different materials (with different moduli), different rod radii, and different radii of bending.</p>
<p>Stated more simply, if I wanted to wind a rod of radius X made of a material with modulus of elasticity M around a spool, how would I calculate the minimum radius S of the spool?</p>
| |applied-mechanics|elastic-modulus| | <p>If we're staying within the realms of beam theory, we can go with this approach which is valid for any material that exhibits linear-elastic behaviour before yield:</p>
<p>The curvature of a beam is related to the applied moment and it's flexural stiffness:</p>
<p>$$
\begin{align}
\kappa = \frac{M}{EI}
\end{align}
$$</p>
<p>where, for a cicular rod, $I = \frac{\pi d^4}{64}$. The curvature is also equal to the inverse of the bent radius, giving the following:</p>
<p>$$
\begin{align}
R = \frac{1}{\kappa} = \frac{EI}{M}
\end{align}
$$</p>
<p>For a rod of circular cross-section, the stress at the extreme fibre is related to the applied bending moment and the elastic section modulus: </p>
<p>$$
\begin{align}
\sigma = \frac{M}{Z}
\end{align}
$$</p>
<p>where for a circular rod, $Z = \frac{\pi d^3}{32}$. This gives:</p>
<p>$$
\begin{align}
\therefore M = \frac{\sigma \pi d^3}{32}
\end{align}
$$</p>
<p>Therefore, if a given material has a certain yield strength, $\sigma = f_y$, the minimum radius can be determined through the substitution of the above equations:</p>
<p>$$
\begin{align}
R_{min} = E \, \frac{\pi d^4}{64} \frac{32}{f_y \pi d^3}
\end{align}
$$</p>
<p>$$
\begin{align}
\boxed{\therefore R_{min} = \frac{E d}{2 f_y}}
\end{align}
$$</p>
<p>where $E$ is the elastic modulus of the material, $d$ is the diameter of the rod and $f_y$ is the yield strength of the material.</p>
<p>Refer to <a href="http://homepage.tudelft.nl/p3r3s/b17_exam_2017_july.pdf" rel="noreferrer">this</a> solution to question 6a on page 8 for a similar solution for a rectangular section. Note that the end result is the same for both a rectangle and circle as both have a distance of $D/2$ from the neutral axis to the extreme fibre.</p>
| 16016 | How to Calculate the Minimum Radius a Rod Can Can Be Bent Staying in the Elastic Range? |
2017-06-26T20:42:10.937 | <p>I have an object that is launched from the side of a plane (it is a flare). </p>
<p>The question is: what does the flare do while it is proximate to the plane? Will it hit the tail?</p>
<p>After accounting for the motion of the aircraft, I have the motion of the flare decently understood and plotted, according to wind resistance without accounting for the propeller wash. </p>
<hr>
<p>Then it occurs that the flare will be in the prop wash of the aircraft while it is in motion near the plane. </p>
<hr>
<p>Understanding that the prop wash will be turbulent, and assuming that I find a way to <a href="https://www.physicsforums.com/threads/how-fast-is-air-blown-from-a-propeller.606908/" rel="nofollow noreferrer">estimate the wind speed of the prop wash</a>, what will be the effect on the force of wind resistance due to the prop wash?</p>
<p>I.e., how will the turbulence of the prop wash mitigate the increased wind resistance due to the airspeed of the prop wash? </p>
<hr>
<p>To expand on this, and the flexibility of the answer, the parameters are: </p>
<ul>
<li>The velocity of the prop wash is an estimate of the upper bound of the <em>average</em> velocity of the prop wash</li>
<li>The purpose of this is to verify that an object coming out of the side of the plane will not hit the tail</li>
<li>It is necessary simply that I establish an upper bound on what the wind resistance conditions are going to be, as I already have a code written for the position estimate, relative to the aircraft geometry, speed, altitude, etc. </li>
</ul>
<hr>
<p>My current solution is as follows (not ready to post as an answer):</p>
<ol>
<li>Get an upper bound on power output of the engine (~1900 kilowatts for the <a href="https://en.wikipedia.org/wiki/General_Electric_T700" rel="nofollow noreferrer">turboprop in question</a> here)</li>
<li>Multiply the watts by the reciprocal of the forward velocity of the aircraft and reciprocal of area affected by the propeller to get a force per sq meter </li>
<li>Solve for $v_w$ in the equation $F = \rho A_{prop} v_o (v_w - v_o)$, which is a correction to the estimate in response 3 of the linked forum thread. </li>
</ol>
| |aerospace-engineering|aerodynamics| | <p>Starting with thrust, </p>
<p><a href="http://nautilus.fis.uc.pt/personal/pvalberto/aulas/cef_mestrado/Air.Resistance.pdf" rel="nofollow noreferrer">http://nautilus.fis.uc.pt/personal/pvalberto/aulas/cef_mestrado/Air.Resistance.pdf</a></p>
<p>$$
T_{upper} = P_{shaft} * \frac{1}{v_o}
$$</p>
<p>Assuming propeller efficiency is 100% (worst case, in the context of this question).</p>
<p>Next, the relation of thrust to propeller wash velocity with respect to the turbo prop in question is described by the following equation (ignoring exhaust): </p>
<p>$$
T_{upper} = Force = \dot{m}(v_w - v_0)
$$</p>
<p>Source: <a href="https://www.grc.nasa.gov/WWW/K-12/airplane/turbprp.html" rel="nofollow noreferrer">https://www.grc.nasa.gov/WWW/K-12/airplane/turbprp.html</a></p>
<p>$\dot{m}$ is only as large as the air which the airplane can reach. Assuming that the velocity remains constant (for a worst case analysis) at whichever thrust level is chosen (presumably, max thrust): </p>
<p>$$
\dot{m} = v_o \rho_{air} A_{prop}
$$</p>
<p>And, solving for $v_w$, </p>
<p>$$
v_w = \frac{F}{\dot{m}} + v_o
$$</p>
| 16041 | Wind resistance inside turbulent flow ("prop wash") |
2017-06-28T18:51:36.607 | <p>I am specifying pipes for a water system with 500 ppm chloride in solution. Ordinarily we would specify 316L stainless steel to limit the corrosion, but I was asked if we could instead specify carbon steel pipe with a corrosion allowance. So instead of the typical schedule 40 pipe, could I realistically use schedule 80 or 120 etc?</p>
<p>My gut reaction is that you will never really even have a "protective" oxide layer with water continually flowing, so you will just continually corrode the pipes out. All that will be gained with thicker pipe is that it will last longer before failure, but you will have increased corrosion products to remove from the flow stream over time.
I could not really find any good resources talking about chloride effects on carbon steel pipe since the general consensus is that stainless should be used for this application.</p>
| |corrosion| | <p>Oxygen and pH will generally be the most important corrosion factors in aqueous corrosion. If this is a potable water as described ,there is no reason to consider anything but carbon or galvanized steel. If the Cl is present as HCl you have a problem that 316 won't solve ; If you you have temperatures above 150F ,you could have stress corrosion cracking of 316 ( although unlikely below 200 F ).
If you have experience with similar conditions , that would be the most useful. Otherwise figure on 5 mpy ( 0.005" ) as a worst case. </p>
| 16074 | Chloride corrosion of carbon steel pipe |
2017-06-28T22:13:56.503 | <p>I'm planning to build a machine which at some point tows a heavy weight on a rope in the air. Once the weight is heigh enough up in the air I want to keep it there for about an hour or so, before I want to put it down again in a controlled manner.</p>
<p>As far as I know, I can brake a DC motor, but I think this takes energy so I would like to avoid doing that for as long as an hour. I guess there should be other options, such as putting a lock of some sort on the rope, but I have no clue how something like that would work.</p>
<p>Could anyone give me some tips on what a usual/best way of blocking a rope for a long time is?</p>
| |electrical-engineering|motors| | <p>One thing that I did to implement a brake system in one of my projects was to install a 20:1 worm-gear reduction. The worm gear, when fully metal and mounted in the proper housing, can act like a brake when stopped, since the interlocking gear cannot turn the worm, but the worm can turn the gear. For this, you will want the motor attached to the worm and whatever equipment you are moving/rotating/mounting/etc.</p>
<p>I would like to note, that while this DOES work - it can and will cause strain and wear on your gears if the load is very heavy. I would recommend that you get a larger gear (more expensive) for heavier loads.</p>
| 16076 | How to stop DC-motor with heavy weight from moving? |
2017-06-29T12:10:48.270 | <p>This question emerges from Earth Science SE since recently there was asked <a href="https://earthscience.stackexchange.com/questions/10695/why-are-co2-sensors-so-expensive-when-co-sensors-arent">this</a>. There so far no one had the engineering/ measurement device expertise to answer this.</p>
<p>I am aware that these are fundamentally different in the case that the purposes of CO2 analyzer is to accurately monitor the levels of CO2 in the atmosphere probably in making a difference between a few parts per million (ppm) in air (as there is currently around 400ppm in the atmosphere and it is a well mixed gas in the atmosphere). On the other hand CO sensors probably need to be able to tell when the concentration of CO exceeds some predetermined threshold where it is safe for humans to breathe.</p>
<p>I would like to hear the answer on why and what causes the large price difference. My guesses are: complexity of manufacturing of CO2 analyzer, parts or the large scale production of CO sensors. It is probably the sum of many but from an expert on this or an enlightened opinion I would like to hear from this more accurately.</p>
| |measurements|product-engineering| | <p>Part of the answer is that CO is chemically reactive, while CO2 is not.</p>
<p>"Cheap and simple" CO sensors either mimic the reaction of CO with the oxygen-carrying chemicals in the bloodstream and detect the color change during the reaction, or they contain a simple fuel cell which combusts the CO to CO2 and measures the electrical current produced. The fuel cell method is now the most commonly used.</p>
<p>CO2 sensors measure the absorption of infra-red light by the CO2 (i.e. they use the same physics which causes the greenhouse effect). This is an accurate method, but not cheap technology. There are cheaper sensors available which use chemical reactions, but the calibration and accuracy is not very stable and they have a relatively short life (e.g. 2 years).</p>
| 16084 | Why are CO2 analyzers much more expensive that CO detectors? |
2017-06-29T13:15:15.710 | <p>When dealing with powders, metal or otherwise, it is often stated weather the powder grains are spherical or flakes. Is it the grinding process that determines the shape of the powder, or is the shape obtained through some other means?</p>
| |materials| | <p>It depends partly on the properties of the material and partly on the manufacturing process. </p>
<p>High quality metal powders used for sintering and powder metallurgy for example are often produced by gas atomisation. Here molten metal is sprayed through a fine jet in an inert atmosphere and the droplets formed are allowed to solidify. This process naturally produces spherical beads of reasonably uniform and controllable size. Variations on this process may use a centrifuge or various gas and liquid jets or an electric arc to disperse the metal. You also get a similar effect as a side effect of some arc welding processes. </p>
<p>Fumed silica, which is widely used as a desiccant, thickening agent and filler for, paints, resins, adhesives etc. is produced by pyrolysis of quartz sand in a flame. </p>
<p>Brittle materials such as hard metal alloys and many minerals can be pulverised in a ball mill, here the particle size and shape will depend on the mill setup and the fracture characteristics of the material in question. </p>
<p>Powders can also be produced by various chemical and electrolytic precipitation processes. </p>
| 16087 | Particle shape of powders |
2017-06-29T17:59:42.383 | <p>In my lab we have dressing tools for sharpening grinding wheels. The tools are diamonds mounted on a steel shank and the shank fits into a custom holder with a set screw in our machine (pictured).</p>
<p><a href="https://i.stack.imgur.com/Ud22v.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ud22v.jpg" alt="Dressing tool holder with set screw"></a></p>
<p>The hole in the tool holder is rather large so that it can fit a variety of tool shank diameters. We have a tool with a shank diameter much smaller than the hole diameter, so it has this metal ring wrapped around it to increase its diameter so that it fits tightly in the holder hole. The ring has a gap on one side so that it can be tightened on the shaft.</p>
<p><a href="https://i.stack.imgur.com/QMwC4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QMwC4.jpg" alt="Metal part to increase the diameter of the shaft, bottom view"></a>
<a href="https://i.stack.imgur.com/iqIzL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iqIzL.jpg" alt="Metal part to increase the diameter of the shaft, side view"></a></p>
<p>My question is: <strong>What do you call this piece of metal used to increase the diameter of a small shaft?</strong> My first thought was to call it a "split collar" or a "split ring" but that doesn't yield anything similar on a Google search.</p>
<p>The reason I am wondering is that we have other dressing tools that are too small to fit in this holder and I would like to buy more of these metal pieces so that we don't need to keep re-using the one we have. But I can't find it if I don't know what it's called.</p>
| |machining|manufacturing-engineering| | <p>This could be called a <a href="http://www.thefreedictionary.com/Bush+(mechanical)" rel="nofollow noreferrer">bush</a> but as no rotary motion it's probably more correct to call it a sleeve.</p>
| 16097 | Part to "increase" the diameter of a small shaft, what is is called? |
2017-06-30T06:03:35.980 | <p>I'm struggling to calculate with any confidence the allowable shear force for a metric bolt holding two aluminium pieces together.</p>
<p>I only need an approximate figure, since I will allow a very large margin of safety. </p>
<p>Assume that the pieces being joined are sufficiently strong. Assume the bolt is threaded throughout its length. It is acceptable for the thread to be damaged, but no permanent deformation to the bolt core. Assume a single bolt holds the two pieces such that the pieces <em>can</em> freely rotate, but that they won't.</p>
<p>The two pieces being joined are are 5mm and 3mm thick. The bolt is M8 with 1mm thread. Assume the bolt is made of a weaker steel.</p>
<p>What is the value of the shear force for the above?</p>
<p>So far, the most progress I have made is by using <a href="http://www.amesweb.info/Screws/Metric_Bolt_Grades_Strength.aspx" rel="nofollow noreferrer">this online calculator</a>, but I think it's for tension strength. Nevertheless, the "proof load" comes out at 8.82kN for for "property grade" 4.6. I'm guessing the area strength is of similar magnitude.</p>
| |bolting|shear| | <p>The <em>Guide to Design Criteria for Bolted and Riveted Joints</em> 1974 by Fisher and Struik, you can use the approximation:</p>
<p>$$S_{Ult-Shear}\sim 0.62\cdot S_{Ult}$$</p>
<p>and from the Distortion Energy Theory: </p>
<p>$$S_{YieldShear}\sim 0.58\cdot S_{Yield}$$</p>
<p>The shear force to yield the bolt is:</p>
<p>$$F = S_{YieldShear}\cdot A_{bolt}$$</p>
<p>Where $$A_{bolt}$$</p>
<p>is the cross sectional area of the bolt in the shear plane</p>
<p>Since the shear plane is across the threaded portion of the bolt, use the minor diameter for an external M8x1 thread, 6.596mm, to calculate the area.</p>
<p>If you're interested in the force required to break the bolt, simply substitute the Ultimate Shear Strength for Yield Shear Strength.</p>
<p>In all likelihood though, the aluminum will tear long before you get to the point of shearing the bolt.</p>
| 16103 | Maximum shear force of metric steel bolt |
2017-06-30T17:17:31.307 | <p>I'm going through some vibration theory and I need to represent the equations in state space for vibration control.</p>
<p>The main forced general equation of motion is:</p>
<p>$$\mathbf{M{\ddot q}+{\Omega_c}G{\dot q}+{[K_b+K_m-\Omega_c^2K_\Omega]q}=T(t)+F(t)}$$</p>
<ul>
<li>$K_b$: bearing stiffness symmetrical matrix</li>
<li>$K_m$: mess stiffness matrix</li>
<li>$K_\Omega$: diagonal centripetal stiffness matrix</li>
<li>$T$: applied external torque</li>
<li>$F$: static transmission error excitation</li>
<li>$q$: the vector with the traslation and rotational coordinates for the sun,ring and carrier</li>
</ul>
<p>If I'm not mistaken, for this equation the input variables would be the forces $F(t)$ and $T(t)$ plus the $q$ terms, which would form the $B$ matrix and $u$ input vector in state space form.</p>
<p>But for the free vibrations I have the equation:</p>
<p>$$\mathbf{M{\ddot q}+{[K_b+K_m]q}=0}$$</p>
<p>Would the input variables in this case just be equal to zero? and the output variables just $\mathbf{{\ddot q}} $ ? I'm also trying to code this into Matlab and Maple, which can directly get me the state space form of really long equations by just inputting the equation, input variables and output variables.</p>
| |mechanical-engineering|control-engineering|dynamics|vibration|eigenvalue-analysis| | <p>Karlo mentioned to use the state space of $[q,\dot{q}]^T$ this allows you to write the system as a first order differential equation of state space. If $X$ were the state space, you could write $\dot{X} = f(X)$. As far as the work you need to do, you write that</p>
<p>$$
X = \begin{bmatrix} q\\ \dot{q}\end{bmatrix}
$$</p>
<p>the derivative of $\frac{dq}{dt}=\dot{q}$ which is part of X and $\frac{d\dot{q}}{dt}=\ddot{q}$ which can be solved for algebraically in the ODE.</p>
<p>$$
\frac{dX}{dt} = \begin{bmatrix} \dot{q}\\ KM^{-1}q \end{bmatrix}
$$ </p>
<p>for the undamped system. If you intend on using a numerical solver in matlab (maybe <code>ode45</code>) it would look something like this assuming all your system parameters are defined as matrices with similar names and <code>IC</code> is the initial condition and <code>tmin</code> and <code>tmax</code> are time bounds you're interested in for a solution.</p>
<pre><code>dxdt = @(x) [x(2); M\K*x(1)];
[q,t] = ode45(dxdt, [tmin, tmax], IC);
</code></pre>
<p><code>ode45</code> specifically expects this form: where the first parameter is a vector that is the derivative of a coordinate/state space vector. Notice we would have the same code if we instead used something that wasn't second order by notation, but the condition that $\dot{s}=t$ makes it second order.</p>
<p>$$
\frac{dX}{dt} = \frac{d}{dt}\begin{bmatrix} s \\ t \end{bmatrix} = \begin{bmatrix} t \\ KM^{-1}s \end{bmatrix}
$$</p>
<p>Using this technique, we can write any system that is a Nth order differential equation as a first order equation of an Nth order state space. </p>
<p>Edit: After seeing your most recent comment, your state space could be written as:
$$
X = \begin{bmatrix} x\\y\\u \\ \dot{x}\\\dot{y}\\\dot{u}\end{bmatrix}
$$</p>
<p>I think I misunderstood what you meant by using matlab and maple. Your inputs are known\controlled values, so <strong>$F(t)$</strong> and <strong>$T(t)$</strong> along with the system parameters, like <strong>$K$</strong>, outputs are system behavior, like $x,y,u$ and their derivatives, what you don't know upon the formulation of the equation, but you don't need maple to construct the equation for you, it's just the ones you find from the algebra. </p>
<p>Lastly, I somehow didn't notice your controller forces <strong>$T(t)$</strong> and <strong>$F(t)$</strong> and didn't write them into the first parameter of the solver, but those can be written in as well, as they are. It would just be that $\ddot{q}=M^{-1}(Kq-T(t)-F(t))$.</p>
| 16116 | How do I represent equations of motion of a dynamic system in State Space? |
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