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2021-09-06T14:36:32.540 | <p>I am estimating the amount of materials required to make a rammed earth building and I can't find anything related with the foundation of rammed earth building and about two-storeyed rammed earth building. Can you tell me about how the foundations are constructed on rammed earth building and what should be done to make two storeyed rammed earth building?</p>
| |structural-engineering| | <p>I would suggest the safest way is to build the foundation with reinforced concrete and leave dowels out to bond with the rebars of the rammed earth wall.</p>
<p>In the middle east, they have build adobe houses for centuries by combining clay and straw as a wall material. This system has proved to have excellent thermal properties, but not sufficient strength for earthquakes.</p>
<p>They used to mix lime with sand and silt and clay as primitive concrete for the foundation.</p>
<p>There is no code for the rammed earth construction. So every individual needs to hire an engineer to do the calculations and prepare the plans for them. here are some photos that were taken during the construction.</p>
<p><a href="http://www.rammedearth.info/rammed-earth-pictures-page-3/#Pictures" rel="nofollow noreferrer">source</a></p>
<p><a href="https://i.stack.imgur.com/ZjPbP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZjPbP.jpg" alt="foundatn" /></a></p>
| 47002 | Rammed Earth Building |
2021-09-07T13:25:48.620 | <p>I'm currently designing a steel structure for a gym machine. I want to assemble two rectangular tubes, one being able to slide inside the other, for height adjusting. I intend to use a spring ball pop pin to sustain the inner tube.</p>
<p>I have not decided which fit and tolerance value to use yet, however, when looking for different dimensions of rectangular tubes, I only find gauges catalogues. These dimensions don't comply with the fits and tolerance dimensions mentioned in handbooks, so my question:</p>
<p>Can tubes for fits and tolerance be purchased? if yes, where can I find them? If no, how can I get those? At a local workshop? Should they be machined?</p>
| |mechanical-engineering|mechanisms|mechanical|machine-design| | <p>For rectangular or square you are pretty much limited to the manufacturer and his reps. Some distributor may happen to have an inventory containing a size you want, but it would be chance. Round tubing is different because some distributors will be able draw down or expand inventory to produce custom sizes; for a price. Also , with round , some distributers can swage down an end to fit into another round. It would be most practical if you could work with standard products.</p>
| 47015 | Can tubes for fits & tolerance be purchased or only manufactured? |
2021-09-07T14:36:43.910 | <p>I need to analyze a continuous beam of length L and three supports: A (at x=0), B (at x=L/2), and C (at x=L). An applied load P is applied at x=3L/4. Now, at A, the continuous beam is bolted to a bracket (drawn in orange). At C, the continuous beam is bolted to another bracket that is welded to another beam (drawn in green), at its midspan (see the image attached). The beam with the welded bracket is perpendicular to the continuous beam.</p>
<p>If I'm not mistaken, <strong>the bracket acts like a cantilever beam</strong>, and <strong>the applied load causes torsion on the perpendicular beam due to the bracket welded</strong>. I tried finding the reaction forces using the <strong>superposition method for statically indeterminate beams</strong>, described in many MoM books (I consulted Philpot's).</p>
<p>According to this method, when dealing with a beam supported by another beam, a compatibility equation must be used, after determining the deflections at the point of interest. The first step when solving this type of problems consists of eliminating the redundant reaction (let's say A, in my structure) and finding the other reactions (B and C) using static equilibrium equations, since eliminating the redundant reaction leaves a statically determinate beam (released beam). In my case, I find that following these steps don't really apply, because reaction force C is determined once a compatibility equation is used when finding deflections at point C. It is something like a paradox: reaction force A is found by finding reaction C, but reaction C is found by finding reaction A, and so on. How can I solve this issue? Is there another analytical method that tackles these cases? Am I just doing something incorrectly? Am I on the edge of discovery of a new MoM method? (just kidding).<a href="https://i.stack.imgur.com/KfaGN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KfaGN.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering|structural-analysis|solid-mechanics|machine-design| | <p>This answer is to point out the possible reason for the "wrong" result getting from the superposition method.</p>
<ul>
<li>The method of superposition is valid and is desirable for this problem, but your assumptions on the support rigidity could have been wrong, thus causing the mistake.</li>
</ul>
<p>The end of the beam on the right (point C) is sitting on another beam that is deflectable and should be modeled as spring support as shown on the graph below. The strength of the spring constant is influenced by the characteristics and properties of the transverse beam - length, stiffness, support conditions, and deflection at point "C".</p>
<p><a href="https://i.stack.imgur.com/1duWJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1duWJ.png" alt="enter image description here" /></a></p>
<p>The first step in solving this problem is to determine the stiffness of the spring. Let's isolate the transverse beam, and apply a unit load (<span class="math-container">$P = 1$</span>) at point "C". The spring constant is then equal to the unit load divided by the deflection found at "C", <span class="math-container">$K = 1/\delta_{C1}$</span>. From here on, everything is back on the track. (<span class="math-container">$\delta_{C1}$</span> is the deflection at point "C" due to the unit load <span class="math-container">$1$</span>.)</p>
| 47019 | Reaction forces on a three support continuous beam with two of them requiring compatibility equations. Superposition method done wrong! |
2021-09-08T00:37:33.520 | <p>EDIT 1:
firstly, thank you for the comments and the input</p>
<p>In this calculations the panels are already involved from the start as being the source (due to other connections to other systems when not heating the pool). But the most important factor is the rise of 17 Celsius in the range from 04 C to 21 C, taking into account very cold conditions - thus the comment that the sizeable amount of water would not cool down from 21 C to literally above freezing in a single night</p>
<p>EDIT 2:</p>
<p>The pool and temperature characteristics: 20ft x 10ft x5 foot deep water body (1000 cubic ft) with temperatures ranging from 21 Celsius in the last light of the day, so something like 25-26 Celsius during the peak of the heat cycle (sometime in the early afternoon) from whenceforward the heat is only maintained by a part of the panels with the other part providing power elsewhere; whereafter the temperature drops to the said 21 Celsius</p>
<p>I have a question regarding pool heating I was thinking about some time ago and if you feel like, share your thoughts.</p>
<p>For some 20ft x10ft and 5 foot deep pool in california in wintertime and outside, how much solar power converted directly into heat for the pool would you guess could heat the pool?</p>
<p>Heating it during the winter's day so that the water is still "mildly lukewarm" in the morning before re-heating (battery heating overnight would be a separate system and is ignored in the calculation).</p>
<p>I'm having an issue with calculating this because of including evaporation and heat loss during the night. But water is complicated since heat is also lost by evaporating since the change from liquid to gas requires latent heat. Anyway, at least for heating:</p>
<p>I was using the constant of 4.2 kW of energy for an change of 1K in 1L of water in 1 second. Meaning, a cubic metre (about 35 ft3) needs 1.1666 kW to heat up 1K/hour.</p>
<p>This is the tricky part: If there was no heat loss, a 1000 ft3 pool (28.3m3) would heat up 7°C in 7 hours and would need about 8.2kW for that. So, 20kW for a 16.8°C increase during 7 hours.</p>
<p>How much "rated" solar power do I need for the panels to have an average power (or, root mean sq power?) of at least 20kW, that is what I am calculating and into this formula should be included also the heat loss during the day.</p>
<p>Because, a rise of 17°C for this amount of water should surely be sufficient to be more than what the water would lose until the morning? At least this is what I assume at the moment. My guess is this outside pool in wintertime (but Californian winter) with a top protective cover/insulation shouldn't lose more than 10-15°C overnight.</p>
<p>In any case the point is how to incorporate evaporation and heat loss into this equation? What other necessary variables should be established? How much (roughly) is a minimum of 20kW average solar power power output to the heaters when taking into account sunlight variation and atmospherics, during an average "real" winter day (i.e. between 21 Nov and 21 Jan) in California?</p>
<p>The panels have an efficiency parameter, 19.78%. This is connected to some specified standard conditions for measuring the performance.</p>
<p>Standard Test Conditions (STC) of radiance of 1000 W/m2, spectrum AM 1.5 and cell temperature of 25°C with the normal operating solar-cell temperature is obtained under the Test Conditions: 800W/m2, 20°C ambient temperature, AM 1.5 Spectrum, 1m/s wind speed.</p>
<p>So, this efficiency is 20% of what? Of the energy in the sunlight, the total amount which could be possibly collected?</p>
<p>I am interested into how all of this actually and precisely is calculated, but primarily my requirement is to find a rough estimate - but, scientifically correct.</p>
<p>So, the zenith hour of the Californian wintertime sunshine could be converted into heat generated by electricity which is in turn converted with a factor of 0.2 from the sunlight on the panel in that hour. But how to incorporate the waxing and waning of sunshine during the day (cloud cover ignored for simplifying purposes!) into all of this? Is there any sort of data website from where it could be noted how much solar radiance changes in a day of a certain season (21 Nov - 21 Jan) for a certain spot on the planet (considering the different inclinations of the rays)?</p>
<p>Any thoughts much appreciated</p>
<p>Regards</p>
| |thermodynamics|heating-systems|photovoltaics|thermal-insulation| | <p>This is not an answer. I will just put a lot of comments about the question. You clearly have some physics background or at least understanding, so hopefully this following comments will be helpful to clarify and proceed.</p>
<h2>Heating up vs. heat losses</h2>
<p>Heating up the pool and bringing it to a set temperature is the easy (and cheap) part. Maintaining the temperature is the difficult part, the evaporative losses are the most significant part.</p>
<p>In that respect its best to calculate the evaporative losses and calculate the average daily <strong>energy</strong> (<em>not power</em>) that you require. So measure things in kWh.</p>
<h2>power vs Energy</h2>
<p>The following line drew my attention</p>
<blockquote>
<p>I was using the constant of 4.2 kW of energy for an change of 1K in 1L of water in 1 second. Meaning, a cubic metre (about 35 ft3) needs 1.1666 kW to heat up 1K/hour.</p>
</blockquote>
<p>IMHO, this approach will complicate maters. You calculate the average power required to heat one cubic meter in one hour. In that time the water will also have losses. And the thing is that you can't tell what those losses are because, you don't know if that water is going from 10<span class="math-container">$^oC$</span> to 11<span class="math-container">$^oC$</span> or 98<span class="math-container">$^oC$</span> to 99<span class="math-container">$^oC$</span>. The losses in those two scenarios are totally different.</p>
<h2>rise of 17°C should be sufficient</h2>
<blockquote>
<p>Because, a rise of 17°C for this amount of water should surely be sufficient to be more than what the water would lose until the morning? At least this is what I assume at the moment. My guess is this outside pool in wintertime (but Californian winter) with a top protective cover/insulation shouldn't lose more than 10-15°C overnight.</p>
</blockquote>
<p>If you are planning to heat up 17<span class="math-container">$^oC$</span> above the set temperature (e.g. 28<span class="math-container">$^oC$</span>) then you'd have to bring the water to 45<span class="math-container">$^oC$</span>. Then:
a) it <strong>just won't be pleasant to use the pool at 45oC</strong>, and you will probably get that early in the afternoon, when ideally you'd like the pool to be cool.<br />
b) the heat losses are proportional to the temperature difference. So it will very quickly loses the heat that it accumulated.</p>
<p>So the trick <strong>is to provide as much heat as the heat losses</strong>. Don't use the pool as a thermal battery or a boiler.</p>
<h2>Approach I would use</h2>
<ul>
<li>find the heat requirements using a online calculator <a href="https://www.engineeringtoolbox.com/swimming-pool-heating-d_878.html" rel="nofollow noreferrer">e.g.</a>. Express that quantity (<span class="math-container">$Q_{req}$</span>) in <span class="math-container">$kWh$</span>.</li>
<li>Go to <a href="https://re.jrc.ec.europa.eu/pvg_tools/en/#PVP" rel="nofollow noreferrer">PVGIS</a> (or <!--<a href="https://globalsolaratlas.info/map?c=32.249974,-115.307007,7&s=33.201924,-117.279053&m=site" rel="nofollow noreferrer">globalsolaratlas</a>) and find at your location and panel an estimate for the <strong>Specific photovoltaic power output</strong> (PVOUT) which is expressed in
<span class="math-container">$\frac{kWh}{ kWp}$</span>.</li>
</ul>
<p>If you divide:</p>
<p><span class="math-container">$$\frac{Q_{req}}{PVOUT}$$</span></p>
<p>you will get a rough estimation of the required photovoltaic installation kW.</p>
<hr />
<h2>Notes about efficiency</h2>
<p>I am not certain what you are planning to do however in terms of flexibility efficiency and cost, I see the following solution:</p>
<ul>
<li>use solar thermal panels (solar mike's suggestion): cost effective, efficient, not as flexible (heating only during the day)</li>
<li>use PV and a heat pump: expensive, very efficient, very flexible because you power the heat pump from the grid</li>
<li>use only PV and convert the heat: expensive, inefficient, not very flexible.</li>
</ul>
| 47029 | Figuring out power requirements of pool heater elements powered directly by solar panels (panel efficiency & sun,evaporation, insulation, ), etc.) |
2021-09-08T15:13:08.977 | <p>Assume a multistage gear reducer (used in forklifts or bed lifts) like the one presented in the following image. Assume that the motor is connected to shaft 1 and the load on shaft 3.</p>
<p><a href="https://i.stack.imgur.com/6i33y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6i33y.png" alt="enter image description here" /></a></p>
<p>Which gear pair has the highest stress?</p>
<ol>
<li>Is it the slowest moving gear towards the lifting end, or</li>
<li>is the stress equal throughout the mechanism?
An internet search has been fruitless.</li>
</ol>
<p>My intuition is #1. Since the mechanism converts velocity to torque, it has high velocity and low torque on the motor side, but low velocity and high torque on the lifting side. So while the strength of individual gear teeth is the same throughout the mechanism, they're subjected to higher forces on the lifting side, and this is where they'll break first.</p>
<p>I'm been presented with the counter argument that it's simply an energy converter, and the same amount of total energy is present throughout the mechanism. So the stress is the same on the motor side as it is on the lifting side. This feels counterintuitive, but I'm unable to refute it.</p>
<p>TO CLARIFY: The question is about multiple gears in succession, e.g. a series of 5 gears, and whether the output stage is under more stress than the input stage, or whether all teeth experience the same stress throughout. I do readily accept that a simple two gear reducer would have the same stress between the teeth of each gear.</p>
<p>Example would be like this, with motor input on the right and load output on the right, and whether the stress is higher on the output stage teeth (left).</p>
<p><a href="https://i.stack.imgur.com/bnHYm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bnHYm.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering| | <p>As per the other answers, when teeth are uniform width, and are engaged, then pressure and shear-stress should generally be highest on the slowest moving gear.</p>
<p>There are also impact loads as they engage and disengage -- in particular with spur gears, as opposed to helical gears (?! Or so I thought?? the reference below doesn't reflect this). Analyzing this is way beyond my experience, but here's a reference (RoyMech) that has proved reliable in other cases:</p>
<p><a href="https://roymech.org/Useful_Tables/Drive/Gears.html" rel="nofollow noreferrer">https://roymech.org/Useful_Tables/Drive/Gears.html</a></p>
<p>There's a formula for stress in there, with a "K_v" factor accounting for impact loads.</p>
<p>If the overall reduction ratio is high, and the velocity on the motor side is high, and the output torque is low, maybe the impact stress on the input side would be dominant as the OP question suspects. I think it would be unusual, but do the calculation and see!</p>
| 47034 | Highest point of stress in a multistage gear reducer |
2021-09-09T01:22:35.907 | <p>Let's say there is 30W of light pumping into the fiber.</p>
<p>Parameters:
Length of Fiber: 6cm</p>
<p>Input - 30 W</p>
<p>Output -29W</p>
<p>Loss of 1W</p>
<p>The diameter of optical fiber is 0.25mm.</p>
<p>Since the optical fiber is extremely light, using this method to calculate the increase in temperature, <a href="https://sciencing.com/how-8643971-convert-wattage-degrees.html" rel="nofollow noreferrer">https://sciencing.com/how-8643971-convert-wattage-degrees.html</a>
we will get a large increase in temperature?</p>
<p><span class="math-container">$1W \times 180s = 180 joules $</span></p>
<p><strong>Assuming a mass of 1g, for 6cm of fiber</strong></p>
<p><span class="math-container">$180 \div 1 = 180 $</span></p>
<p><strong>Glass with a specific heat of 0.792J/g.C</strong></p>
<p><span class="math-container">$180 \div 0.792 =227.37^\circ $</span></p>
<p>But in real life, optical fibers do not get heated up, to this temperature, approximately 32 to 60 degrees Celcius.</p>
<p>I tried doing a simulation, with parameters:</p>
<p>Emissivity of Fiber(Glass) : 0.92</p>
<p>Convection Coefficient of <span class="math-container">$10W/m^2 K $</span></p>
<p>And an internal Heat of 1W.</p>
<p><a href="https://i.stack.imgur.com/9NQ3S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9NQ3S.png" alt="enter image description here" /></a></p>
<p>This is inconsistent with real life, ignoring other losses for fiber such as splicing and connection loss, why is this so?</p>
| |optics|photonics|fiber-optics| | <h2>Your assumption on temperature increase is flawed</h2>
<p>You can't convert watts to temperature directly unless the object is perfectly insulated and the application of heat has a limited time. The equation from your link would result in a temperature increase directly proportional to time, increasing forever.</p>
<p>Temperature is a measure of "heat content," which changes based on heat_in minus heat_out. As soon as the fiber begins to heat up it will pass heat out from itself to the environment. The engineering work that has to be done is to determine the heat going out from the fiber.</p>
| 47042 | Calculating Temperature of an Optical Fiber |
2021-09-09T03:00:24.540 | <p>Hi in my engineering class we are currently making bridges and our group decided to make a parabolic bridge that will hopefully support 600 newtons (point load).</p>
<p>the rules are that the bridge must span 0.4 meters (40 cm) and can be held together by bolts and made out of popsicle sticks.</p>
<p>currently, we have an equation of the parabola which is y=-a(x+20)(x-20) but a, in this case, is a mystery to us.</p>
<p>my question is what value of (a) makes the parabola the strongest.</p>
<p>thank you so much and please help</p>
| |bridges| | <p>This is to help you to find the equation of a parabola.</p>
<p><a href="https://i.stack.imgur.com/7XCaE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7XCaE.png" alt="enter image description here" /></a></p>
<p>In order to create a parabola, we need to set a few points on it to define the curve (see graph below), and to set up the equation. For example, let's say h = 10, b = 20, and the form of the equation is <span class="math-container">$y = f(x) = -ax^2 + C$</span> (a & C are constants).</p>
<ul>
<li>Solving the constant "C":</li>
</ul>
<p>At the apex, <span class="math-container">$(x,y) = (0,10), y = f(0) = -a(0)^2 + C = 10$</span>, so <span class="math-container">$C = 10.$</span></p>
<ul>
<li>Solving the constant "a":</li>
</ul>
<p>At the left end point, <span class="math-container">$(x,y) = (-20,0), y = f(-20) = -a(-20)^2 + 10 = 0$</span>, solving, <span class="math-container">$a = 0.025$</span>.</p>
<p>So the equation of this parabola is <span class="math-container">$y = -0.025x^2 + 10$</span>, or <span class="math-container">$y = \dfrac{x^2}{40}+10$</span>. As noted before, the higher the curve, the more efficient in load-carrying capacity. However, for the bridge deck, you need to take into account the driveability too.</p>
| 47043 | how to calculate the best possible equation of a parabola to hold the most point load |
2021-09-09T21:56:16.417 | <p>Part is welded to a plate, based on the below free body diagram how would I determine the max stress in the weld? It seems half the weld is in tension and the other half is in compression.</p>
<p><a href="https://i.stack.imgur.com/bvlD1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bvlD1.png" alt="free body diagram " /></a></p>
<p><a href="https://i.stack.imgur.com/bvlD1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bvlD1.png" alt="enter image description here" /></a></p>
| |structural-engineering|mechanical| | <p><strong>Sum Loads About Centroid of the Welds</strong></p>
<p><a href="https://i.stack.imgur.com/EJjED.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EJjED.png" alt="enter image description here" /></a></p>
<p><strong>Max. Weld Stresses:</strong></p>
<p><span class="math-container">$f_{c,t} = \dfrac{P}{A} \pm \dfrac{M}{S}$</span> (lbs/in, along each weld length)</p>
<p><strong>Alternative:</strong></p>
<p>If the above weld pattern does not produce enough strength, before increasing the weld size, try to switch the welds to other sides as shown below. Other than the increase in section modulus, everything remains the same.</p>
<p><a href="https://i.stack.imgur.com/vrJ18.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vrJ18.png" alt="enter image description here" /></a></p>
<p><strong>Note: The square tubing usually has round corners, which will result in a shorter effective weld length.</strong></p>
<p><a href="https://i.stack.imgur.com/PxTo3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PxTo3.png" alt="enter image description here" /></a></p>
| 47066 | Max Stress on Weld |
2021-09-10T06:42:00.077 | <p>Recently I am building equipment. It is as large as two refrigerators with aluminum extrusion frame. Unfortunately, it has visible resonance vibration. After measuring I find on the floor there is a 16.5 Hz vibration, probably from elevator motor above floor, or ventilation motors about 20 meters away.</p>
<p>So, I want to change my design, enhance rigidity to increase resonance frequency. Customers may install it near a normal vibration source. But, what are the low frequency of building vibration? I need to know it first to avoid it. Does anyone know the range of building low frequency? Usually from the elevator, transformer, ventilation motor, or water pump. Or is there any general frequency standard based on experience that a product resonance frequency should be higher than it to avoid resonance vibration?</p>
| |mechanical-engineering|vibration| | <p>As an equipment supplier, your responsibility is to avoid the resonance to occur between the equipment and its support, whether a skid or a frame that is to be supplied by you. It might be a problem down the road, but it's out of your hand to evaluate the fundamental frequency of an industrial building, which may have numerous vibratory equipment and excitation sources.</p>
<p>You shall clearly indicate the vibration characteristics (operation frequencies) of your equipment including its support, and the associated parameters (mass, geometries) on the vendor drawing though.</p>
<p>For troubleshooting the vibration problems in a building, the owner shall engage a lab or engineering consultant to perform the measuring and analysis.</p>
| 47075 | What frequency is the low frequency vibration of normal buildings from like elevator or ventilation motor? To avoid equipment resonance vibration |
2021-09-10T16:38:35.923 | <p>In strength of materials, there are two types of axial forces which can act on a prismatic bar - Compressive and Tensile. Do they significance only in pairs? <a href="https://i.stack.imgur.com/R9e4V.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R9e4V.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-analysis|applied-mechanics| | <p>If a compressive force is applied, the bar will only experience compressive forces. If a tensile force is applied, the bar will only experience tensile force. They can only occur simultaneously on the same bar if the bar is subjected to bending.</p>
<p>The picture you have shared, it can be one of these cases. 1) Either you are pulling (tensile) or pushing (compressive) both the ends of the bars 2) You are fixing one end and pulling or pushing the other end 3) You are making an imaginary perpendicular cut somewhere in the bar and then observing this picture.</p>
<p>Lets consider case 1 now; both of your hands need to be putting a force in opposite direction to push or pull the bar. In case 2, the fixed end is doing the same thing as your other hand was doing in case 1. In case 3, you still observe the cutted cross section of the bar having an equal but opposite force (as compare to what you are applying to it). This can also be thought of as Newton's third law analogy for simplicity.</p>
<p><em><strong>It should be noted that the top two pictures that you have shared will only be possible when the bar is static and not in a rigid body motion. Since you haven't shown any reaction forces in the bottom two pictures, it might do a rigid body motion (assuming the contact with the ground is frictionless).</strong></em></p>
| 47086 | Do tensile and compressive forces have significance only in pairs? |
2021-09-11T06:23:46.020 | <p>I tried to model a problem months ago and still I can't solve, I don't want forget this problem so I would like to share what I tried and if you know what is happening with my graph.</p>
<p>This is the problem is related with the world record free fall:</p>
<p>Felix Baumgartner (m = 75 kg) recently stepped out of a balloon at an elevation of 40,000 m and parachuted to earth, setting a height record for skydivers. Here we simulate his jump from the time he steps out of the balloon until the time he opens the parachute. Assume the drag force FD is of the form:</p>
<p><span class="math-container">$ F_D = -k|Vy|Vy $</span></p>
<p>where <span class="math-container">$k$</span> is the drag coefficient and <span class="math-container">$ V_y $</span> is the vertical velocity. The effect of elevation on the gravitational constant is minor. However, because air density depends on elevation and he jumped from a very high elevation, the effect of elevation on the drag coefficient k should be accounted for. A crude model for k that takes into account the variation of air density with elevation is</p>
<p><span class="math-container">$ k = k_{0}e^{-0.00011y} $</span></p>
<p>where <span class="math-container">$k$</span> is the drag coefficient at elevation <span class="math-container">$y$</span>, <span class="math-container">$k_0$</span> is the drag coefficient at sea level <span class="math-container">$(y = O)$</span> and where <span class="math-container">$y$</span> is in <span class="math-container">$m$</span>, and <span class="math-container">$k$</span> and <span class="math-container">$k_0$</span> are in <span class="math-container">$kg/m$</span>. Assume <span class="math-container">$k_0 = 0.16 kg/m$</span> before the parachute opens.</p>
<p>a. Using a coordinate system where <span class="math-container">$y$</span> is positive upward, write the differential equations that represent the velocity <span class="math-container">$v_y$</span> and elevation <span class="math-container">$y$</span> of Felix during the time between stepping out of the balloon and opening his chute. Include the initial conditions.</p>
<p>b.Simulate and graph Felix's velocity and elevation as functions of time. At
what elevation will he have reached his maximum velocity? What is his
maximum velocity? For comparison, measurements during the jump indicated that his maximum velocity was <span class="math-container">$1358 km/h$</span> and his free fall from <span class="math-container">$40,000$</span> to <span class="math-container">$2000m$</span> (when his chute opened) lasted for <span class="math-container">$260s$</span>.</p>
<p>I tried this:
<span class="math-container">$ a $</span></p>
<p>Initial Conditions: <span class="math-container">$ V(0) = 0\frac{m}{s} $</span></p>
<p><span class="math-container">$ FD = -k|Vy|Vy $</span></p>
<p><span class="math-container">$ FD = -kV^2 $</span></p>
<p>Before the parachute opens:</p>
<p><span class="math-container">$ F = ma $</span></p>
<p><span class="math-container">$ F = ma $</span></p>
<p><span class="math-container">$ F = m\frac{dv}{dt} $</span></p>
<p><span class="math-container">$ W + FD = ma $</span></p>
<p><span class="math-container">$ W = mg $</span></p>
<p>System model:</p>
<p><span class="math-container">$m\frac{dv}{dt} = -kv^2 - mg $</span></p>
<p><span class="math-container">$\frac{dv}{dt} = \frac{-kv^2}{m} - g $</span></p>
<p><span class="math-container">$\frac{dv}{dt} = \frac{-0.16e^{-0.00011y} v^2}{m} - g $</span></p>
<p><span class="math-container">$a = \ddot{y} , v = \dot{y} , y=y.$</span></p>
<p>2nd order differential equation:</p>
<p><span class="math-container">$ \ddot{y}= \frac{-0.16e^{-0.00011(y-y0)} \dot{y}^2}{m} - g $</span></p>
<p>This is my block diagram:</p>
<p><a href="https://i.stack.imgur.com/6ZuiX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ZuiX.png" alt="enter image description here" /></a></p>
<p>This graphic show my error:</p>
<p><a href="https://i.stack.imgur.com/VeW7m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VeW7m.png" alt="enter image description here" /></a></p>
<p>What I am forgetting?</p>
<p>Bibliography:</p>
<p>A First Course in Differential Equations, Modeling, and Simulation - Carlos A. Smith, Scott W. Campbell</p>
| |control-theory|modeling|matlab|experimental-physics|systems-engineering| | <p>Although this is a solution in Python (not in Matlab)</p>
<p>It produced the following graphs (which make sense due to the increased aerodynamic friction in the lower layers of the atmoshpere.</p>
<p><a href="https://i.stack.imgur.com/4K2gR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4K2gR.png" alt="enter image description here" /></a></p>
<p>Additionally the code below produces also the phase portrait which is:</p>
<p><a href="https://i.stack.imgur.com/M2LvQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M2LvQ.png" alt="enter image description here" /></a></p>
<p>the following code creates the above:</p>
<pre><code>#%%
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt
k=0.16 # kg/m
g = 9.81 # m/s^2
y0 =40000 # m
m=75 # kg
def baugartner(X, t):
y = X[0]
vy = X[1]
dydt = vy
dy2dt = 1/m *(k*np.exp(-0.0001*y)*vy*vy) -g
return [dydt, dy2dt]
X0 = [y0, 0]
t = np.linspace(0, 300, 2500)
sol = odeint(baugartner, X0, t)
y = sol[:, 0]
vy = sol[:, 1]
fig, axs = plt.subplots(2,1, sharex=True)
axs[0].plot(t,y, label='height [m]')
axs[1].plot( t, vy, label='velocity [m/s]' )
plt.xlabel('t')
axs[0].set_ylabel('height [m]')
axs[1].set_ylabel('velocity [m/s] ')
plt.legend(('y', 'v_y'))
# # phase portrait
plt.figure()
plt.plot(y,vy)
plt.plot(y[0], vy[0], 'ro')
plt.xlabel('height')
plt.ylabel('velocity')
# %%
</code></pre>
| 47101 | Unstable System. Error Trying to model the Red Bull Stratosphere Free Fall |
2021-09-11T14:28:08.463 | <p>In a Winchester 1873, the carrier block is the part that moves a cartridge from the magazine up to the chamber, and that pushes the previously spent shell up out of the top of the rifle. It moves back down and out of the way before the cartridge is fired.</p>
<p>The carrier block seems to always be made entirely out of brass, and it seems to be the only part of the rifle that is made out of brass.</p>
<p>Why is this? I figure that brass sinks a lot of heat and is relatively elastic, but neither of those properties seem relevant since the block isn’t near the cartridge when it is fired. Why is it not made out of steel or another cheaper metal?</p>
| |heat|firearms|brass| | <p>Brass is a natural choice for accomodating sliding or rotating contact with steel surfaces. This is because oxide grit or metal particles which would cause galling get embedded into the (softer) brass instead of peeling material off of both sliding surfaces, as commonly happens when the sliding surfaces are made of the same material (bad design choice!). This is particularly important if the sliding surfaces are not well-lubricated and exposed to dirt- both conditions being met in gun mechanisms.</p>
| 47106 | Why is the carrier block in a winchester rifle made of brass? |
2021-09-11T18:14:48.610 | <p>I am using the aluminum design manual equations to design an aluminum column that fixed at the base and free at the top, so fixed-free end conditions. Section C.3 of the Aluminum Design Manual says to use an effective length factor of k=1 for all members. The effective length factor for fixed-free columns is 2.It seems like I cannot use the equations in the manual and will have to use theoretical equations (eulers/secant). When using k=2 and the design manual equations my allowable stress was extremely low.</p>
| |mechanical-engineering|structural-engineering|columns| | <p>You should use <span class="math-container">$ K=2 $</span></p>
<p>and the allowable stress</p>
<p><span class="math-container">$$\sigma_{allowable}= \frac{\pi^2*E}{n_u(\frac{kl}{r}^2)} \quad (Eq. 3.4.7-3) $$</span></p>
<ul>
<li><p><span class="math-container">$n_u= buildings safety factor =1.9 $</span></p>
</li>
<li><p><span class="math-container">$r= radius of \ gyration=\sqrt {I/A}$</span></p>
</li>
</ul>
<p>Source: Aluminum Design Manual 2005.</p>
<p><a href="https://i.stack.imgur.com/aa8nX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aa8nX.png" alt="safety factors" /></a></p>
| 47111 | Clarification on Aluminum Design Manual |
2021-09-11T22:18:30.833 | <p>When working on a technical drawing project, I couldn't find the exact dimensions of the following screw</p>
<p><a href="https://i.stack.imgur.com/1aBjX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1aBjX.png" alt="enter image description here" /></a></p>
<p>In particular, I don't understand the <span class="math-container">$0.5:1$</span> to <span class="math-container">$\emptyset 15$</span> indication.</p>
<p>Is this screw in any DIN/ISO (maybe DIN 914 or ISO 4027) standard?</p>
| |technical-drawing| | <p>Expanded from Solar Mike's answer:</p>
<p>The screw is 8mm in diameter, 22mm long, has a thread every 1.25mm, and the tip is 5mm in diameter, tapered down from the full screw diameter (8mm) with 0.5(H): 1(V) tapering.</p>
| 47115 | What are the exact dimensions of this screw? |
2021-09-12T10:05:32.653 | <p><strong>Edit 3</strong>: the question being closed due to being opinion based? The design causes problems that are not in the other design, there are no benefits provided for the restrictions that the design imposes, therefor it is an inferior/wrong design. I was really hoping to see any benefits for the traded off usability but seems there are none.</p>
<p>The chargers with USB port on a right or left hand side on a power board intrude on the space of the next outlet next to them, the charger with USB port on top doesn't have such a problem, my question is what is the point of symmetry breaking chargers?
There are at least 3 other options that are not intrusive on the next outlet on a power board.
Here is one from a big enough manufacturer that should have known better about the problems of this design by now (2021).</p>
<p><strong>Edit1</strong> : Is there a patent issue that might be forcing a manufacturer to prefer this flawed design over the other? while browsing google images I noticed more side types from the same manufacturer.</p>
<p>Besides patent constraints to opt for a worse design are there other any other possible factors?</p>
<p><strong>Edit 2:</strong> Thanks to perfect answer by jsotola , it seems the charger might have been designed for a different setting and was never tested for the other settings, even though it might have been a perfect solution at one point having it moved over to other markets and settings no reviews was made and a perfect solution became a flawed design.</p>
<p>The only other possible explanation besides patent constraints on design could be financial and lack of a QA review process for a design.</p>
<p><a href="https://i.stack.imgur.com/lOQp0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lOQp0.png" alt="enter image description here" /></a></p>
| |design|power-electronics| | <p>I think that the power adapter was probably first designed for the North American market.</p>
<p>For that market, the design is not flawed.</p>
<p><a href="https://i.stack.imgur.com/GAfp7m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GAfp7m.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/E1vocm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/E1vocm.png" alt="enter image description here" /></a></p>
| 47125 | Why design chargers with USB port on a side rather than top |
2021-09-12T16:22:43.617 | <p>As a layman, how can I know the shear strength of a particular size bolt?</p>
<p>For example, if I want to attach something to the internal wall of my van ( 1mm stainless steel ) using a rivnut and standard metric bolt and it should hold up to 50 kg, how would I figure out the right size and grade to use? Nothing specialty, just standard stuff a layman could get at a typical hardware store.</p>
<p>In my specific case I'm installing wooden cabinets on the internal walls of my Fiat Ducato van, but I'd like as general an answer as reasonable.</p>
| |shear| | <p>I will start by highlighting the main points (some were also found in other answers). I will additionally provide a short calculation for the <em>tensile strength</em> (and <em>shear strength</em>) very quickly by knowing the <strong>type</strong> (e.g. M6) and <strong>grade</strong> (e.g 10.9). and below I will providing a layman's calculation for a metric bold.</p>
<p>In the numerical calculation I will use a very small, and very low grade bolt to show case that the bolts usually aren't the weak link in this case.</p>
<h2>the important parts for a layman</h2>
<ul>
<li>(tiger guy's): <strong>Bolts shouldn't be used in shear</strong> (they can be used but in most
scenarios are not suggested)</li>
<li>for 50 kg any metric bolt should be able to hold.</li>
<li>(Daniel K) In many cases, bolts are not the weak link in the assembly.</li>
</ul>
<h2>Calculation for bolt tensile strength</h2>
<p>The <strong>type</strong> M6 is <em>indicative</em> of the diameter of the core. So the cross-sectional diameter (d)of the core is (<em>close to</em>) <span class="math-container">$A = \frac{\pi \cdot d^2}{4}=\frac{\pi\cdot 6^2}{4}$</span></p>
<p>The <strong>grade</strong> shows the material type. It consist of two numbers separated by a dot (<strong>X.Y</strong>). E.g. 8.8.</p>
<ul>
<li>The first number multiplied by 100 is the UTS of the material MPa</li>
<li>The second number (with the dot) is the ratio of the yield stress to the UTS of the material.</li>
</ul>
<p>So for an 5.6 bolt the UTS is <span class="math-container">$5\cdot 100 =500 MPa$</span>, and the yield stress is <span class="math-container">$0.6\cdot \sigma_{UTS}=300 MPa$</span>. The followonig table has some of the most common bolt grades.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;"></th>
<th style="text-align: center;">UTS</th>
<th style="text-align: center;">yield stress</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">4.6</td>
<td style="text-align: center;">400 MPa</td>
<td style="text-align: center;">240 MPa]</td>
</tr>
<tr>
<td style="text-align: center;">5.6</td>
<td style="text-align: center;">500 MPa</td>
<td style="text-align: center;">300 MPa</td>
</tr>
<tr>
<td style="text-align: center;">8.8</td>
<td style="text-align: center;">800 Mpa</td>
<td style="text-align: center;">640 MPa</td>
</tr>
<tr>
<td style="text-align: center;">10.9</td>
<td style="text-align: center;">1000 MPa</td>
<td style="text-align: center;">800 MPa</td>
</tr>
<tr>
<td style="text-align: center;">etc</td>
<td style="text-align: center;">..</td>
<td style="text-align: center;">..</td>
</tr>
</tbody>
</table>
</div>
<p>So the <em>tensile strength</em> up to yield strength of an M4 4.6 is:
<span class="math-container">$$ A \cdot \sigma_y = \frac{\pi\cdot (4 [mm])^2}{4} \cdot 240 \left[\frac{N}{mm^2}\right]= 3015 N$$</span></p>
<p>Or approximately 300 kg.</p>
<h2>Calculation for bolt shear strength</h2>
<p>For the shear strength, you can approximate the shear yield stress to 0.5 of the tensile yield stress (this is a <strong>very</strong> conservative approximation if you compare it to tables from Eurocode). So the shear load should be in the order of (at least) 150 kg.</p>
<hr />
<p>Please note that I used in the example one of the smaller metric bolts M4 (its hard to find a wrench them), at the lowest grade, and it still was capable of holding (At least) 3 times the 50 kg load.</p>
| 47128 | Metric bolt shear strength for a layman |
2021-09-13T11:23:20.733 | <p>I've noticed that prescription eye glasses frames are often made from metal that is quite rigid. It's surprisingly difficult to bend the metal, despite how thin it is. And if it does bend, it doesn't seem to weaken much when bent back and forth.</p>
<p>Is there a certain kind of metal that is typically used for making prescription glasses frames?</p>
<p>Examples that come to mind:</p>
<ul>
<li>High-grade aluminum or aluminum alloy</li>
<li>Stainless steel (can it be painted?)</li>
<li>Titanium</li>
</ul>
| |materials|metals| | <p>There are many different materials that can be used for modern eyewear.</p>
<p>Based on the following table (I put it together from online searches, its not from a single site, although the content is repeated - also I haven't managed to complete it), probably the ones you are looking at are either:</p>
<ul>
<li>Berryllium alloy</li>
<li>some memory alloy (usually titanium based Ti e.g. Flexon)</li>
</ul>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;">description</th>
<th style="text-align: center;">lightweight (<span class="math-container">$gr/cm^3$</span>)</th>
<th style="text-align: center;">strong</th>
<th style="text-align: center;">corrosion resistant</th>
<th style="text-align: center;">hypoallergic</th>
<th style="text-align: center;">cost</th>
<th style="text-align: center;">flexible</th>
<th style="text-align: center;">usage</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">Titanium</td>
<td style="text-align: center;">+++ (4.5)</td>
<td style="text-align: center;">+++</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
<td style="text-align: center;">durability</td>
</tr>
<tr>
<td style="text-align: center;">Monel (usually nickel-copper alloys)</td>
<td style="text-align: center;">(8.8)</td>
<td style="text-align: center;"></td>
<td style="text-align: center;">++</td>
<td style="text-align: center;">+ (have a coating to protect the skin)</td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
<td style="text-align: center;">people who spend a lot of time in or around salt water</td>
</tr>
<tr>
<td style="text-align: center;">Beryllium alloy</td>
<td style="text-align: center;">++ (1.85+?)</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;">+++</td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
<td style="text-align: center;">++</td>
<td style="text-align: center;">people who spend a lot of time in or around salt water</td>
</tr>
<tr>
<td style="text-align: center;">Stainless steel</td>
<td style="text-align: center;">+ (8)</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
<td style="text-align: center;">++</td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
</tr>
<tr>
<td style="text-align: center;">Flexon (Ti -Ni shape memory alloy)</td>
<td style="text-align: center;">+ (6.5)</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;"></td>
<td style="text-align: center;">+++</td>
<td style="text-align: center;">active kids</td>
</tr>
<tr>
<td style="text-align: center;">Beta titanium (Ti+ traces Al+V)</td>
<td style="text-align: center;">++ (4.5)</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;"></td>
<td style="text-align: center;">+++</td>
<td style="text-align: center;">active kids</td>
</tr>
<tr>
<td style="text-align: center;">Aluminum</td>
<td style="text-align: center;">+ (2.7)</td>
<td style="text-align: center;">++</td>
<td style="text-align: center;">++</td>
<td style="text-align: center;">+</td>
<td style="text-align: center;"></td>
<td style="text-align: center;">++</td>
<td style="text-align: center;">high-end frames</td>
</tr>
<tr>
<td style="text-align: center;">Magnesium</td>
<td style="text-align: center;">+ (1.7)</td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
<td style="text-align: center;"></td>
<td style="text-align: center;">high-end frames</td>
</tr>
</tbody>
</table>
</div>
<p><em>NOTE:</em> the above is not a authoritative table. It is my interpretation from online searches.</p>
| 47141 | What kind of metal are eye glasses frames made from? |
2021-09-13T17:02:26.650 | <p>When we load a body, say a prismatic bar, axially, internal resistive forces are developed within it to oppose the elongation or compression. What was the need of relating these internal resistive forces with area to define a quantity like stress. Wasn't these internal resistive forces enough?</p>
| |mechanical-engineering|structural-engineering|materials|structural-analysis| | <p>First courses on engineering statics tend to start from the forces where the structure is loaded and constrained, and at the interface between components, because in simple situations (for example statically determinate structures) the forces can be calculated independently of the internal stresses in the structure. The stresses can then by found (approximately) from the forces.</p>
<p>However in general the stress and strain distribution is are more important. As a simple example, consider a rectangular plate of material with tensile forces applied to each end. You can easily calculate the <em>average</em> tensile stress as "force/area".</p>
<p>However, consider what happens if there is a <em>hole</em> cut in the bar for some reason. Assume there are no forces applied around the circumference of the hole.</p>
<p>It should be obvious that the stress in the material near the hole will change in some way, and the change might be important. In fact, for a single small hole, the maximum stress is increased to <em>3 times</em> the average stress. For a larger hole comparable with the width of the bar, or for several holes close together, the stress pattern is even more complicated.</p>
<p>Another example is a rectangular section cantilever beam. You can calculate the <em>approximate</em> stress distribution in the structure by finding the applied loads and moments and then, <em>making some assumptions about the deformed shape of the beam</em>.</p>
<p>For many purposes those assumptions are good enough, but in fact they are not even self-consistent. For example, suppose you know the material has non-zero Poisson's ratio (because you have measured it, not from some theoretical argument!). When the beam bends, you assume there is an axial tensile stress on one surface and a compressive stress on the opposite surface.</p>
<p>Therefore, because of Poisson's ratio, the <em>width</em> of the beam on the surface with tensile stress will <em>decrease</em> and the width on the opposite surface will increase. That means the beam section is no longer a rectangle, and in fact it is <em>curved</em> from one side to the other.</p>
<p>Google for "anticlastic curvature* for pictures - for example <a href="https://www.researchgate.net/figure/The-phenomenon-of-anticlastic-curvature-arises-from-Poisson-effects-When-a-beam-is-bent_fig4_11057987" rel="nofollow noreferrer">https://www.researchgate.net/figure/The-phenomenon-of-anticlastic-curvature-arises-from-Poisson-effects-When-a-beam-is-bent_fig4_11057987</a></p>
<p>Again because of Poisson's ratio, the <em>volume</em> of the material in tension is different from the volume in compression. Therefore the neutral axis of the beam is <em>no longer at its geometrical center</em> when the beam is bent.</p>
<p>Also, because of the shear stress through the depth of the beam, the assumption that a plane section remains plane is also false, and in fact the original plane section deforms into an S-shape viewed from the side.</p>
<p>None of this could be calculated just by considering the <em>forces</em> applied to the beam and then working out the "average" stress in the material.</p>
| 47146 | Why we needed to relate internal resistive forces with area (stress)? |
2021-09-15T04:42:13.403 | <p>I have to manufacture a tracked robot weighing nearly 40 Kg's and the operation is mostly on rusted metal surface.</p>
<p>I will be using 2 rubber(1 in*30 in)tracks for movement but I'm worried it will wear sooner than tires has anyone worked on this, so that I know how long I can operate the robot until the tire completely wears.</p>
| |robotics| | <p>These factors come to mind more prominantly</p>
<ul>
<li><p>The rusty surface could have chemicals that change the rubber and make it brittle and thus shorten its life.</p>
</li>
<li><p>The track has to be tight and not too wide to give it manouverability and proprtioned not to lag behind the wheels due to its inertia and cuase spining of the wheel</p>
</li>
<li><p>The tracks have to be rather short and have small indentations and be chamfered at the edges to promote flexibilty and traction and turn on command.</p>
</li>
</ul>
| 47167 | Rubber tracks for robots |
2021-09-15T16:32:56.437 | <p>In the smart glasses the images are projected on glass. But they are still clear to eyes.</p>
<p>We usually can't see something clearly at such close distance. Then, how is the image projected from a small screen on the glass?</p>
<p>Can someone explain how it works in actual smart glasses also how it can be possible if I want to project a screen(not so small, small oled ones) on a glass making it visible to close distance?</p>
| |optics|glass| | <p>When you project an image on a reflective medium, you inherit the projection distance. So what you are seing is a plane that is hovering behind the glass. In essence your eyes see is the whole path to the target as if it never had reflected at all.</p>
<p>So no need to focus that close.</p>
| 47173 | How can a small screen be projected on glass but still visible near 2cm from eyes? |
2021-09-16T02:16:43.943 | <p>I've searched on Google that aluminium foil can be a thermal insulator, but how does it work?</p>
<p>From my thought process, it should've been a conductor because its a metal, right? But, some said that because its so thin and has large surface area, the heat would be transferred to its environment quickly, hence will cool it.</p>
<p>If I wrap something with an aluminium foil, will it cool that thing that is wrapped? I'm thinking it wouldn't. Yes, it is logical that because its thin and large surface area it will cool it by dissipating the heat quickly, but to where? I'm thinking if I wrap it, it might dissipate it to the inside as well, transferring the heat to the wrapped object also.</p>
<p>Is my thought process wrong? If yes, where? I really need some explanation to this. Thank you very much!!</p>
<p><strong>EDIT</strong></p>
<p>To anyone who needs more context: So I have this sensor with operating temperature of 70 degree Celsius, and the measured ambient temperature is nearly that value, so I'm thinking that I need to wrap it with something to keep it cool. I have aluminium foil on hand, will it work if I wrap the foil around it to keep it cool? I'm thinking that it will eventually get hot, but will it be cooler in longer duration if I wrap it?</p>
| |heat-transfer|aluminum| | <p>One thing that hasn't been mentioned in other answers is the extent to which shiny pale bodies radiate less than 'black bodies'</p>
<p>So, if you take e.g. a jacket potato, and wrap it in tin foil, then you will get several benefits:</p>
<ol>
<li>The foil reflects some IR back in</li>
<li>There is a small insulative air gap</li>
<li>The heat which enters the foil is quickly conducted to the outside surface of the foil, but, it then radiates away into to the air much more slowly than it would have done from the rough dark surface of the potato would have radiated directly.</li>
</ol>
| 47189 | Can aluminium foil be a thermal insulator? |
2021-09-16T08:43:47.287 | <p>The tensile test specimen used for tension test has an enlarged cross sectional area at the ends. The sources I'm referring to, suggest that this is the case so that the failure occurs in the central region of the specimen where the stress distribution is uniform and is more easier to calculate (using P/A).</p>
<p>Now, what I conclude is that if the c/s area of the specimen were to be uniform throughout the length, failure would've occured near the ends where the stress distribution was non uniform, and that is the reason why we increase the area near the ends so that failure doesn't occur near the ends. My question is if the c/s area were to be uniform throughout the length why the failure occurs near the ends?</p>
<p>I mean if we are enlarging the ends, we would've been doing so because failure is occuring near the ends (with uniform c/s), so we're trying to avoid that by enlarging the ends.</p>
<p>If what I have written isn't making sense, then if anyone could explain what would've been the problems faced with a uniform c/s area specimen, that would be enough too.</p>
| |mechanical-engineering|structural-engineering|civil-engineering|structural-analysis| | <p>One does not want the rupture in the grips; that makes elongation and reduction of area measurements very difficult. Also, notch sensitive materials will rupture in the grip area unless the the grips are significantly larger cross-section than the gage length. For notch sensitive materials even the shoulders must have a long gradual taper to avoid shoulder ruptures.</p>
| 47192 | Why tensile test specimen is enlarged at its ends? |
2021-09-16T11:02:31.643 | <p>Let's say I have a rod that I need rotating, like in a regular crank mechanism:</p>
<p><a href="https://i.stack.imgur.com/ysIaY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ysIaY.png" alt="enter image description here" /></a></p>
<p>Now, the motor I use is too weak to generate the required amount of torque to move the mechanism. So I put a gear with 10 teeth on the motor, and replace the wheel with a 40-toothed gear. Now I have a mechanical advantage of 4, so the driven gear can deliver 4 times the amount of torque that the motor can.</p>
<p>Does the amount of torque that the driven gear can transfer differ when I connect the rod close to the hub of the gear, compared to when I connect it closer to the rim?</p>
| |gears|torque| | <p>l
Let's say the radius of your gear is R, and you connect the rod at distance x from the center, with the gear's torque 20Nm.</p>
<p>The force your rod receives from the gears is</p>
<p><span class="math-container">$$ F_{rod}=\tau*\frac{R}{x}=20\frac{R}{x}$$</span></p>
<p>The smaller the x, the distance from the center the bigger F. But its displacement is smaller.</p>
<p>It works like a lever with a fulcrum at the center of the hub which you try to lift an object at a distance of x from the fulcrum. and you have an action force</p>
<p><span class="math-container">$$F_{gear}=\frac{\tau}{R}$$</span></p>
<p><span class="math-container">$$\frac{F_{rod}}{F_{gear}}=\frac{R}{x}$$</span></p>
| 47195 | Does the amount of torque that can be transferred from wheel to rod change based on the distance of the connection to the rim of the wheel |
2021-09-16T16:15:31.357 | <p>My background drafting is structural bridge drawings. I had Inventor LT installed the other day and I have been playing with it to refamiliarize myself with it and get used to the differences between it and AutoCAD for drawing. I found a pdf book of examples that I had been going through in AutoCAD and decided to carry on in Inventor LT. I am stuck on the following example:</p>
<p><a href="https://i.stack.imgur.com/aTY2e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aTY2e.png" alt="Example 77" /></a></p>
<p>I started by drawing the cross section of the rim and central hub as sketch1. I then revolved them around the central axis. I also realize I could have made a circle and extruded the central hub as another approach.</p>
<p><a href="https://i.stack.imgur.com/WI086.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WI086.png" alt="Sketch1" /></a></p>
<p>For sketch2 I drew up a view looking along the axis of the wheel.</p>
<p><a href="https://i.stack.imgur.com/1tBnY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1tBnY.png" alt="Sketch2" /></a></p>
<p>I had three stumbling blocks in completing this.</p>
<ol>
<li><p>The first stumbling block was the position of the R10 rib around the
outside center of the center hub. The only way I have been able to
position this so far would be to make it to the R13 fillet/arc
between the spokes at the central hub. If this assumption is
correct how do I make a plane at this point to draw the R10 circular
rib?</p>
</li>
<li><p>The second stumbling block I have is drawing the spokes in 3D. My
intent here was to draw an ellipse 22X9.5 and 24X11 at the point
where the R6 and R13 curves/fillets start/end and then lofting from
one to the other. Similar to my first stumbling block, how do I
create sketch planes at these specific spots?</p>
</li>
<li><p>My third stumbling block is how to extrude/join the spoke from the
ellipse to the hub or rim while respecting the R6 and R13
curves/fillets?</p>
</li>
</ol>
<p>This is where I am currently at with the overall drawing:</p>
<p><a href="https://i.stack.imgur.com/Wf8cD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wf8cD.png" alt="3D Model" /></a></p>
<p>Software Version: Autodesk Inventor LT 2021</p>
<h1>Update</h1>
<p>I figured out how to place a working plane through the R13 curve and the center of the wheel. I was then able to project the geometry of the apex of the R10 rib around the central hub. I drew a portion of a circle and on the sketch plane, and then revolved it around the central access.</p>
<p>Using this new ability for me to place sketch planes I then proceeded to produce a sketch plane at the start and end of the spoke where it is 22 nd 24 on the major axis and 9.5 and 11 on the minor axis.</p>
<p>I tried extruding with rails but that just generated errors. I also tried extruding the end of the spoke to the next object but that also gave errors. If it is actually possible to extrude the face of an existing solid, will it extrude based on its existing tapper or will it just extrude straight?</p>
<p>Current State:</p>
<p><a href="https://i.stack.imgur.com/EEAfV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EEAfV.png" alt="Floating spokes" /></a></p>
<p><a href="https://i.stack.imgur.com/FjHNj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FjHNj.png" alt="enter image description here" /></a></p>
| |technical-drawing|drawings|autodesk-inventor| | <p>tl;dr - add the fillets at the end.</p>
<p>Step 1: Revolve. This should also include the R10 'lump' on the central axis.</p>
<p><a href="https://i.stack.imgur.com/s5n3im.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s5n3im.png" alt="Step 1" /></a></p>
<p>Step 2: Define the Elipse Taper. It's stated that the thickness is 11mm and 9.5mm respectively at each end of the spoke. I don't want to start this spoke at the surface of the hub as then my loft would need extending in order to fully intersect it. Instead, I will loft from the XZ plane. So, I need to know how big to make the elipse there! Create reference sketches, with the dimensioned values (11mm and 9.5mm in this view, and 22mm and 24mm in the other axis of the spoke).</p>
<p><a href="https://i.stack.imgur.com/gtMDrm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gtMDrm.png" alt="Step 2" /></a></p>
<p>Step 3: Loft. I created a plane by selecting the bottom point of my reference sketch, and the XZ plane before clicking plane. This automatically creates a plane that is parallel to the XZ plane and passes through that point. The two ellipses are easy to create by first projecting the endpoints of the reference sketches, and then making these coincident to the major and minor axes of the elipse.</p>
<p><a href="https://i.stack.imgur.com/pTrO8m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pTrO8m.png" alt="Step 3" /></a></p>
<p>Step 4: Pattern. Pattern the loft feature around the Z axis</p>
<p><a href="https://i.stack.imgur.com/GlTMXm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GlTMXm.png" alt="Step 4" /></a></p>
<p>Step 5: Cut the bore and keyway. I also champhered here since I didn't include it in the original sketch.</p>
<p><a href="https://i.stack.imgur.com/bP62Lm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bP62Lm.png" alt="Step 5" /></a></p>
<p>Step 6: Add the 3mm Fillets to the roots of the ribs (could also be in original sketch if you prefer), and then add the 6mm and 13mm fillets to the roots of the spokes</p>
<p><a href="https://i.stack.imgur.com/K2HGIm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K2HGIm.png" alt="Step 6" /></a></p>
<p>Step 7: Admire your work.</p>
<p><a href="https://i.stack.imgur.com/7wY8g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7wY8g.png" alt="Step 7" /></a></p>
| 47205 | How to place a sketch plane at a specific geometric location |
2021-09-16T19:05:25.777 | <p>I'm new to mechanical engineering.</p>
<p>I have a 5m long lever with a fulcrum positioned 1m from one end. The whole setup is balanced and I need to calculate the torque of a motor that would move the lever around.</p>
<p>My biggest concern is the inertia of the whole rig, as I need it to do precision movements.</p>
<p>Can you please help me?</p>
<p>rough sketch:</p>
<p><a href="https://i.stack.imgur.com/FqsoX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FqsoX.jpg" alt="enter image description here" /></a></p>
| |motors| | <p>Assuming that:</p>
<ul>
<li>the counterweight and load are point masses,</li>
<li>the load mass is <span class="math-container">$m_L$</span></li>
<li>the bar has mass <span class="math-container">$m_b$</span></li>
</ul>
<p>Then the counterbalance mass should be equal to
<span class="math-container">$$m_{cw} = \frac{3 m_b + 8 m_L}{2}$$</span></p>
<p>The moment of inertia for this problem is:</p>
<p><span class="math-container">$$I_{total} = I_{counterweight} + I_{bar} + I_{load}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$I_{cw}$</span> is the counterweight mass moment of inertia
<span class="math-container">$$I_{cw} = m_{cw} L_{cw}^2 = \frac{3 m_b + 8 m_L}{2} [kg.m^2]$$</span></li>
<li><span class="math-container">$I_{load}$</span> is the load mass moment of inertia
<span class="math-container">$$I_{L} = m_{L} L_{L}^2 = 16\cdot m_L [kg.m^2] $$</span></li>
<li><span class="math-container">$I_{bar}$</span> is the bar mass moment of inertia</li>
</ul>
<p><span class="math-container">$$I_{L} = \frac{m_{b} L_{b}^2}{12} + m_{b}\cdot (\frac{3}{2})^2 = \frac{13}{3} \cdot m_b [kg.m^2] $$</span></p>
<p>Therefore (if I done all the algebra correct):</p>
<p><span class="math-container">$$I_{tot} = \frac{35}{6} m_b + 20\cdot m_L [kg.m^2]$$</span></p>
| 47209 | How to size a motor for a balanced lever? |
2021-09-16T20:56:46.417 | <p>I have a laptop power adapter shown in the picture below.
<a href="https://i.stack.imgur.com/Zo4Ml.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zo4Ml.jpg" alt="enter image description here" /></a></p>
<p>It is written there</p>
<p>170Watt</p>
<p>INPUT: 100-240V ~ 2.5A</p>
<p>OUTPUT: 20V 8.5A</p>
<p>I believe for AC Input they mean Root mean square voltage/current.</p>
<p>for output it is clear (<span class="math-container">$W=U \cdot I$</span>) or <span class="math-container">$170W = 20V \cdot 8.5A$</span>.</p>
<p>But I do not understand it for input. If it is working for minimal input voltage 100V, then minimum power should be <span class="math-container">$100V \cdot 2.5A = 250W$</span> or for 240V <span class="math-container">$240V \cdot 2.5A = 600W$</span>.</p>
<p>Even for minimum 100V power is 250W, then where the difference between input and output 250W-170W=80W goes? I mean from energy conservation law output power should be equal to input power minus losses or <span class="math-container">$Output Power = Input Power \times Energy Conversion Efficiency$</span></p>
<p>Does that mean my adapter has 68% efficiency on 100V and 28% on 240V?
I believe Energy Conversion Efficiency for modern AC/DC adapters should be quite high around 80-90% or I am wrong? This adapter is quite new. I bought it in January 2020 and it was a new one and a new Laptop model.</p>
<p>My question is: Why there is so big difference in input and output power or what is wrong with my calculations?</p>
| |electrical-engineering|power-electronics|electrical|current|ac| | <p><a href="https://www.lenovo.com/us/en/p/accessories-and-software/chargers-and-batteries/chargers/4x20e50574" rel="nofollow noreferrer">Lenovo ADL170NLC3A</a> has an Energy Star V (ES 2.0) rating. <a href="https://www.energystar.gov/ia/partners/prod_development/revisions/downloads/eps_spec_v2.pdf" rel="nofollow noreferrer">Energy Star</a> states to get that rating for nameplate output power (<span class="math-container">$P_{no}$</span>) > 45W, the Minimum Average Efficiency in Active Mode ≥ 87%.</p>
<p>At nominal conditions of 170W, 20V @ 8.5A, and a minimum efficiency of 87% implies 195W input. So worst case, the ac charger consumes 25W.</p>
<p>The AC adapter is a switching power supply converting 100V to 240V ac at 50Hz or 60Hz to 20V DC @ 8.5A. The 2.5A corresponds to the peak ac draw, not sustainded draw.</p>
<p>Typically, the actual current drawn by the laptop depends on the charge on the battery. This means efficiency changes. When the battery is uncharged, output from charger will be maximum and efficiency will be at a low of 87%. As battery charges, demand on charger will be less as battery voltage increases. Efficiency should increase slightly because the 25W lost to switching power supply will change slower than load from battery. Educated guess, efficiency > 90%.</p>
| 47216 | AC/DC Laptop adapter power losses |
2021-09-17T08:00:29.177 | <p>I initially put this one on the physics-side, but this might be a more appropriate place:
<a href="https://physics.stackexchange.com/questions/666116/mollier-diagram-and-reading-the-relative-humidity-when-knowing-the-dry-and-wet">https://physics.stackexchange.com/questions/666116/mollier-diagram-and-reading-the-relative-humidity-when-knowing-the-dry-and-wet</a></p>
<p>I had a thought-error when reading Mollier diagrams and thankfully we have the internet to assist, but even after finding the correct sources showing the answers, I can't get out of my thought-loop to understand why my thinking is wrong. I hope someone could in simple and very definite terms explain why my thinking is wrong.</p>
<p>So let's say that I have the dry bulb temperature of 20 Celsius and wet bulb temperature of 15 Celsius and I'd like to know the relative humidity.</p>
<p><a href="https://i.stack.imgur.com/Umg0j.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Umg0j.jpg" alt="enter image description here" /></a></p>
<p>My thought error was that I thought that I'd read this from the 100% humidity point at 15 Celsius, straight up along the absolute humidity lines. My thought here was that the absolute humidity is the same and the air's water content doesn't change no matter how much I spin my wet-bulb thermometer. So I'd have assumed from the image for the relative humidity to be about ~72% at 20 C. (red arrow)</p>
<p>However this source showed how it should be read and I was also able to test it online. So I should actually read it along the enthalpy lines, which gives me about 58% relative humidity at 20 C. (yellow arrow)</p>
<pre><code>https://www.engineeringtoolbox.com/humidity-measurement-d_561.html
http://www.ringbell.co.uk/info/humid.htm
</code></pre>
<p>Now I'm a bit stuck: I can from one perspective understand that the enthalpy does not change and is constant, but I'm having a hard time to definitely understand why my initial assumption about the absolute humidity is the "wrong constant" in this case.</p>
<p>Could someone take it down to my level, and also possibly clarify if there is something I still seem to have misunderstood, or what should be good to recognize?</p>
| |temperature| | <p>Absolute humidity isn't the same if the dry bulb temperature changes. The definition of absolute humidity is the actual amount of water vapor in the air, while relative humidity is the ratio between the water vapor and the amount of water that the air can hold (saturation) at that given dry bulb temperature.</p>
<p>If the air gets hotter (dry bulb increases) the amount of water that the air can hold increases. <strong>While the amount of water in air remains the same (absolute humidity stays), the amount of water vapor that the air can hold changes, and thus changing the relative humidity.</strong> However, at some point <strong>if the air becomes saturated due to the change of the amount of water that air can hold, the absolute humidity will also change as some of the water vapor will condense.</strong></p>
<p>If you go straight up with the red arrow you are implying that the hotter the air gets, the absolute humidity of the air remains the same. While <strong>in reality, at some point the absolute humidity changes when the air is saturated and some of the water vapor condenses.</strong></p>
| 47228 | Mollier diagram and reading the relative humidity when knowing the dry- and wet bulb temperatures |
2021-09-18T12:38:29.230 | <p>Can anyone please tell me the topics I need to study from <strong>Fluid Mechanics</strong>, specially fluid dynamics to understand <strong>Heat Transfer</strong> in a better way?</p>
| |fluid-mechanics|heat-transfer| | <p>Fluid mechanics is mostly used in handling convection problem, whether it is forced convection or natural convection. Additionally, on those two problems the concept that is used is mostly dimensionless number (like Re, Gr, Pr, etc.) and that's it, and also boundary layer concept. If there is one concept that you should try to understand, I think try to learn what those dimensionless number tell you about how the fluid behave.</p>
| 47250 | Pre requisites for a course in Heat Transfer from Fluid Mechanics , for a Mechanical Engineering Bachelors Degree |
2021-09-18T16:14:59.287 | <p>I am getting two ways to find F. One is Fx0.2=30 Nm which gives F=150N Other is Fx0.2-640x0.13=0 which gives F=416N I am confused which one is incorrect. I know problem is too basic but still not clear where I am wrong. Question is shown in picture. PS: Sorry for such picture, don't have better one.</p>
<p><a href="https://i.stack.imgur.com/Kyn1G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Kyn1G.jpg" alt="Question" /></a></p>
| |pressure|applied-mechanics|moments|pistons|braking| | <p>My interpretation is "the force causing the rod to pressure the hydraulic cylinder that results in 640 N compressive force in the rod/cylinder, and a net moment/torque of 30 Nm about the pivot point. If so, the force should be:</p>
<p><span class="math-container">$0.2F + 30 - 640*0.13 = 0$</span></p>
<p><span class="math-container">$F = 266 N$</span></p>
<p>Other than the 30 Nm moment, the pivot point has a horizontal force that is equal to <span class="math-container">$640 - F_x$</span>, and a vertical force that is equal to <span class="math-container">$F_y$</span>. <span class="math-container">$F_x$</span> & <span class="math-container">$F_y$</span> are component forces of <span class="math-container">$F$</span> in <span class="math-container">$x$</span> & <span class="math-container">$y$</span> axes respectively.</p>
<p>Note, the 640 N force was drawn as the applied force in the direction of action. For equilibrium consideration, its direction needs to be reversed.</p>
<p><a href="https://i.stack.imgur.com/XWYuj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XWYuj.png" alt="enter image description here" /></a></p>
| 47253 | What will be the force F for the given brake system? |
2021-09-18T17:21:27.487 | <p>I have a 6" x 6" x 1/4" piece of carbon filled teflon that is slightly curved (I think it must have come off an roll of enormous diameter) that I need to flatten. The carbon fill is carbon black, not carbon fiber, in order to make the material static dissipative.</p>
<p>Is there any good procedure to flatten it like with an iron?</p>
<p>Alternatively, thickness is not too important so in theory I could try stoning it to flatten. I will have to at some point to remove scratches on the surface but I imagine it will glaze the stone like nothing else and I don't want to need to stone too much of it since this material produces nasty dust.</p>
<p>With a chord of 6", there is a distance of about 2mm to the circumference so it would turn the 6mm plate into a 2-3mm plate if I were to stone it down to flatten it, which is fine if it were not for all the work involved and dust that would be produced.</p>
<p>UPDATE: Stoning does nothing. Just messes up the stone. But wet sanding with Scotch-Brite 7447 (Maroon) cleans up scratches and imperfections real nicely. A circular motion produces a very nice matte finish. Finer grades would probably produce a finer finish. Doesn't flatten it though.</p>
<p>UPDATE: Glass transition temperature is 120C so I think I'll try to heat it up beyond that. This is going to be used to make really thick 2" diameter washers. So I think I will cut out oversized square pieces, mount them on a big bolt with 2" steel washers separating them, maybe with a spring to continuously apply force and then put the whole thing in an oven at 150C or so...if I can get the oven to get that low.</p>
| |materials| | <p>I have some experience dealing with off the shelf PTFE 'curves' and thought I would provide the technique we used to straighten our material as it may be helpful for your project.</p>
<p>For this particular case we were dealing with tubing that came off of a roll, likely much in the same as your material, and it caused it to have a natural curve. Every attempt at bending didn't last as the material had 'memory' and given enough time would go back to the original shape, regardless of how many times it was straightened.</p>
<p>Picture of the tube as shipped:</p>
<p><a href="https://i.stack.imgur.com/2HPvr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2HPvr.png" alt="enter image description here" /></a></p>
<p>This was problematic as we were using it in the Central Thimble of a research reactor for reactivity core measurements and this bend caused clearance issues when passing through the s-bend on this particular port (as one could imagine, it is not okay to force items through a position on a nuclear reactor).</p>
<p>There weren't many material options due to issues with activation from radiation and a certain level of materials toughness was required under operating conditions, so this had to be made to work at the time.</p>
<p>Our best approach was putting together a simple jig to over-bend the tube in the opposite direction while applying heat with an off the shelf heat gun. This worked quite well once we figured out the correct temperature for our teflon polymer.</p>
<p>As PTFE is a flourocarbon it is quite resistant to applied heat with a melting temperature typically above 320C so can handle quite a bit of abuse, although these materials require fairly high temperatures before they will 'hold' the new shape. An infared thermometer is useful, but this could be done without if need be. In either case, be careful and wear appropriate PPE including heat resistant gloves or you could easily get burned.</p>
<p>It is easy to over-correct the natural bend in the opposite direction at temperature so it is best to practice and be willing to settle with 'close enough' as we did on our rig:</p>
<p><a href="https://i.stack.imgur.com/o5OUa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/o5OUa.jpg" alt="enter image description here" /></a></p>
<p>Nearly a year later, and after a number of irradiations in the core, the apparatus is generally in the same shape as when we first straightened it.</p>
<p>As your material is a sheet it should be easier to manipulate as you will have less challenges with potential collapse and the carbon likely will not cause much, if any, additional issues.</p>
<p>Hope this helps!</p>
| 47256 | Flattening teflon plate |
2021-09-19T02:59:22.120 | <p>I am curious to know if the St. Louis Gateway Arch would be strong enough to act as a support structure for a giant vertical axis wind turbine.</p>
<p>I way I am envisioning this being constructed is that a radial shaft bearing block could be installed just below the top of the Arch and a combined radial & thrust bearing block could be constructed down at ground level. 600 feet of steel piping would be installed between the top bearing block and the ground level bearing block. The turbine blades would be attached to this steel piping and the rotating steel piping could be connected to an electric generator on the ground via some sort of pulley and chain setup.</p>
<p>Could the St. Louis Gateway Arch be used as a support structure for a giant vertical axis wind turbine?</p>
<p><strong>EDIT</strong></p>
<p>Here is a drawing of what a vertical axis wind turbine might look like positioned underneath the Arch:</p>
<p><a href="https://i.stack.imgur.com/Ldvnu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ldvnu.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering|turbines|wind-power| | <p>The arch has been designed to resist wind loads but just the wind load its 630 feet arch can cause.</p>
<p>The motto of not over designing the structure to reduce the weight of the structure to a manageable weight has lead to even reducing the viewing openings on top to 7 by 27 inches.</p>
<blockquote>
<p>Over 500 tons of pressure was used to jack the legs of the Arch apart for the last four-foot piece to be inserted at the top. A larger window would not withstand that pressure. from park official guide.</p>
</blockquote>
<p>Adding a huge vertical windmill between the base of the arch and its top will impart lateral loads by orders of magnitude above what the arch is been designed for and cause a catastrophic collapse.</p>
<p>It also will destroy the grace and beauty of the arch that has been designed with a sense of drama and poetry to reflect the river and moonlight.</p>
| 47260 | Could the St. Louis Gateway Arch be used as a support structure for a giant vertical axis wind turbine? |
2021-09-19T19:41:16.300 | <p><a href="https://i.stack.imgur.com/ngCp1.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ngCp1.png" alt="enter image description here" /></a></p>
<p>Image Source: <a href="https://mechtics.com/design/strength-of-material/stress-strain-curve-for-mild-steel-with-defination-of-stress-and-strain/" rel="noreferrer">https://mechtics.com/design/strength-of-material/stress-strain-curve-for-mild-steel-with-defination-of-stress-and-strain/</a></p>
<p>I know that A is the proportional limit, B is the elastic limit. Up to B if the loading is removed the material will restore its original dimensions in theory. C is the upper yield point, beyond C plastic deformations occur.</p>
<p>I have referred many books but couldn't find what's happening in the portion BC? the material is behaving elastically or plastically?</p>
| |mechanical-engineering|structural-engineering|civil-engineering|steel| | <p><strong>Summary: There is increasing plastic behavior and decreasing elastic behavior as the sample is strained from B to C. The transition is caused by random grain orientations and variable resolved shear stress.</strong></p>
<p>With respect, the other answers generally have the right idea, but are missing the important meso-scale mechanism for why there is a non-linear shape between points B and C.</p>
<p>Between points B and C on the stress-strain diagram there is a transition from elastic stretch to plastic flow. The transition occurs because bulk metals are typically composed of many randomly-oriented crystal grains. As you may know, metal crystal grains flow by application of shear stress. The resolved shear stresses on the crystal grain slip systems is what matters for initiating plastic flow. Grains at different orientations experience different resolved shear stresses. Grains more "favorably" oriented, with larger resolved stresses, experience plastic flow earlier. Because grain orientations are random, each grain will start to flow at a specific stress level, giving a distribution of stresses from point B to C. At the macro-scale the distribution means that the sample gradually transitions from "no plastic flow" at point B to "all flow" at C.</p>
<p>At least, until unlocking of interstitial atoms trapped at dislocations starts to occur between points C and D. See <a href="https://engineering.stackexchange.com/questions/13583/why-does-the-stress-strain-curve-decrease/13603#13603">this QA</a> for reasons why D is at lower stress than C, and for why NMech's <a href="https://engineering.stackexchange.com/a/47281/421">answer</a> plot is shaped the way it is.</p>
<p>For a little discussion of resolved shear stresses see <a href="https://engineering.stackexchange.com/q/8407/421">this QA</a>.</p>
| 47271 | What happens from B to C in this stress strain diagram of mild steel? |
2021-09-21T19:46:30.583 | <p>Can someone please explain what I am losing/gaining in the trade-off between Volts and Amp-Hours when buying a piece of equipment such as a chainsaw.</p>
<p>I'm currently comparing two items. I'm making this concrete with actual products because I think this will draw attention to what we (the layman) are struggling with. These are both Battery cordless chainsaws</p>
<ul>
<li>DeWalt 20V Lithium-Ion with 5.0 Ah Battery</li>
<li>Ryobi 40V with 4.0 Ah Battery</li>
</ul>
<p>As pointed out in the comments those are Amp Hours and not Amps. So let's pivot this question: How does someone with almost zero knowledge of battery chainsaws work out what Voltage is appropriate for their usage?</p>
<p>Or maybe a better question is: What do I need to understand about Volts in this context to make a better decision on a purchase like this based on just that information? i.e. if everything else was equal, how do Volts enter into the equation based on the application?</p>
| |electrical-engineering| | <p>I will try to add something Tiger Guy's answer, which I think is spot on.</p>
<h2>Ah and V</h2>
<p>For the DeWalt 20V Lithium-Ion with 5.0 Ah Battery, the 5Ah Battery means that is can provide energy equal to 1A for 5h (or 5A for 1h). However you don't know the actual discharge rate. So, you only know the capacity of the battery (same as knowing the size of the fuel tank of a vehicle).</p>
<p>If you multiply Volts with Amps you get <strong>Wh</strong>. So</p>
<ul>
<li>DeWalt : 100 Wh</li>
<li>Ryobi : 160 Wh</li>
</ul>
<p>From the comparison the Ryobi has a larger capacity (so I would expect the batteries to be bulkier).</p>
<h2>Voltage and motors.</h2>
<p>This is the tricky part. Regarding what people in general refer to as a powerful motor/hand tool is (IMHO) the <strong>torque</strong> that it produces. I.e. if a tool produces more torque then it is more "powerful".</p>
<p>The torque of a motor is strongly related to the motor (and more specifically its characteristics) and the current flowing through it. Typically nowadays (I hope I am not that outdated), power tools use brushless DC Motors.</p>
<p><a href="https://i.stack.imgur.com/cxRkQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cxRkQ.jpg" alt="enter image description here" /></a></p>
<p><strong>Figure: Brushless DC Motor (Source: <a href="https://www.youtube.com/watch?v=bCEiOnuODac" rel="nofollow noreferrer">youtube video</a>)</strong></p>
<p>However, (in most DC motors) - for a given configuration (e.g. number of windings, thickness and resistance of wire) if you increase the voltage, the current also increases and therefore you increase the torque. Brushless motors in particular have a <strong>KV</strong> rating which shows the rpm per voltage applied.</p>
<p>However, the configuration of the motor is not fixed, so you can have different motors at each product, with different power requirements. Unless you actually go into the datasheet of the power tool and maybe even the motor, its not easy to understand which is more powerful.</p>
<h2>Final thoughts</h2>
<p>(I don't expect this will help you directly with your decision which chainsaw to buy, but) keep in mind that <strong>numbers next to technical quantities are often used (almost) out of context for marketing purposes from companies</strong>. So, (usually) you get what you pay for.</p>
<p>Also, I'll close with continue the analogy from vehicle: The Battery is the fuel tank and the motor is ... the vehicle engine (and the drivetrain to some extent). You can have a more powerful engine which means that you can go faster and accelerate more (but the fuel consumption is greater so your run out of fuel faster. Or with a smaller engine, the acceleration and the speed is less, but the fuel tank lasts more <em>hours</em>. However, unless you pop the hood, or start up the vehicle, you don't really know what engine or the drivetrain is underneath.</p>
| 47305 | Amp-Hours vs Volts in electric chainsaw specifications |
2021-09-22T07:41:34.360 | <p><a href="https://i.stack.imgur.com/40K78.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/40K78.jpg" alt="enter image description here" /></a></p>
<p>A: Proportional limit| B: Elastic limit| C: Upper YP | D: Lower YP</p>
<p>Let us say, the gage length was 100mm (random, just for the sake of understanding). I know that if I unload the body anywhere between OB (O is the origin), the body will return back to its original shape and size. So if in the region OB it got to a length of say 102mm while loading, after unloading the body will again come back to 100mm.</p>
<p>I have two questions,</p>
<ol>
<li><p>If suppose, I stretch the body up to anywhere in the region BC to a length of 104mm (say), then while unloading the body, the body wont come back to 100mm but to somewhere say 103mm. (some elastic recovery will take place). I have seen this in one of the videos on Youtube but I want to make sure if this is right?</p>
</li>
<li><p>If suppose I stretch the body up to anywhere in the region DE to say 110mm, then on unloading will the body remain at 110mm itself or will it show some recovery?</p>
</li>
</ol>
| |mechanical-engineering|structural-engineering|civil-engineering|structures| | <p>Generally what happens is that if when the load is removed (quasi statically) the material will return in a path which is parallel to the elastic region.</p>
<p>Eg. for load up to E and F the following behaviour is observed.</p>
<p><a href="https://i.stack.imgur.com/liWrc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/liWrc.jpg" alt="enter image description here" /></a></p>
<p>Notice that the slope of the red arrows is identical to the slope of OA section (region up to proportionality point).</p>
<h2>hysteresis</h2>
<p>If the load is not removed quasi-statically (fancy word for very slowly), then the path that the material returns to the point will change. (see black lines).</p>
<p><a href="https://i.stack.imgur.com/b7AGN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b7AGN.png" alt="enter image description here" /></a></p>
<p>This is because the internal damping of the material has an additional stiffening effect.</p>
<p>Despite the change in the path, the final position that the material eventually returns to should remain the same (assuming you are not doing this in elevated temperatures etc).</p>
| 47313 | What happens when we remove the load, in different regions of the stress strain diagram of mild steel? |
2021-09-22T13:36:32.283 | <p>I'm always confused on what is the effective way to work on CAD with multiple team members to make it efficient. So far, as I'm freshly graduated I only work on a project where I'm the only one using the CAD software. At best there's only two of us, so we can work in one table and discuss directly to each other to be aligned on the part dimensions.</p>
<p>That is what I'm confused, how do team members align all the parts if they work with a lot of team member?</p>
<p>From what I know, on big projects there could be hundreds or even thousands of part, and I'm sure it is not only made by one person. But, if we make CAD in team, the problem that I come through in my experience is: I need to really align on how to assembly the component that they make to the component that I make. What makes it hard is it's really hard to be aligned on what's the boundary of each part, whether or not if the part is made in certain dimension it will hit other part. This is even harder when there is a mechanism, if it moves it will potentially hit other part as well</p>
<p>I'm also confused on how to divide the work in CAD projects. Say there is part A and B, which will be assembled together. If these part is done by different person, wouldn't it be a risk that it cannot be assembled properly? In this case its just two parts, what if there is like 50 or even 100 parts?</p>
<p>So basically, my question is, how is these works done in the industry? Especially for the big projects where there's lots of parts to be designed. I just can't figure out how to do this properly in team and hence I prefer to work alone.</p>
| |mechanical-engineering|design|manufacturing-engineering|cad|project-management| | <p>I have no experience in the engineering industry, but I used to be part of an undergraduate engineering competition team that designed and built planetary rovers and autonomous underwater vehicles. Perhaps I can provide a perspective of doing CAD in a small team of 10 to 20 students.</p>
<ul>
<li>Members of the team usually have an idea of how the final product might look like. This might be based on their experience from building the previous iteration of a design, from building prototypes, from reading trade magazines, or from reading competitors' competition papers from previous years. Members meet, discuss, and make a rough sketch of the final product with rough dimensions. In addition, the first draft of a todo list is written. Using the rough sketch and the todo list, the team can start using CAD to produce the first iteration of the design.</li>
<li>The project is divided into subteams of 2-5 people, with each subteam responsible for a different component. For example, there is a hull team responsible for hull design, a propulsion team responsible for the evaluation/acquisition/integration of propulsion systems, an electrical team responsible for power systems and sensors, etc. Each subteam maintains its own todo list. Each subteam eventually produces CAD designs of the parts that it is responsible for.
<ul>
<li>Normally, each part is input into CAD by only one or two people, after their subteam decides that it is able to do so. For example, when the electrical subteam has decided to use a particular off-the-shelf power system, the subteam will then create a CAD part of that power system, and modify any other parts that integrate with that power system.</li>
<li>Every part is placed into a version control system such Autodesk Vault. This ensures that two people are not modifying the same part at the same time. In addition, this allows the team to keep track of history and easily revert to a previous design if needed.</li>
<li>The use of parameters in parametric CAD allows dimensions to be changed easily when it is time to combine one part with another part.</li>
</ul>
</li>
<li>During weekly whole-team meetings, the project is discussed at a high level. This includes monitoring progress, modifying the todo list, discussions about integration between parts (i.e. the fixing of dimensions between parts), etc.</li>
<li>Not all subteams will complete their designs at the same time, but after many iterations of the above, the parts will start to approach a point where they can be seamlessly fitted into an assembly.</li>
</ul>
| 47321 | How to work on big CAD project in team? |
2021-09-22T15:22:47.313 | <p>I'm trying to interpret the following mechanical drawing:</p>
<p><a href="https://i.stack.imgur.com/2heNv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2heNv.png" alt="enter image description here" /></a></p>
<p>However, I struggle with the "tolerances" part. Shouldn't one dimension (like <span class="math-container">$44$</span> diameter) have one number attached to it, to indicate that tolerance? In this case we have two numbers, and I can't understand their meaning.</p>
| |technical-drawing| | <p>Generally drawings that have tolerances simply listed and not specifically attached to a dimension are general tolerances that apply to <em>all</em> dimensions, note this would not apply for the thread, as this has it‘s own tolerances.</p>
<p>In this case it‘s a Dimension oversized between .220 and 0.42</p>
<p>Agreeably however this is a poorly made drawing, but as it seems to be a simple set ring you‘d use to prevent movement on an axel, it would make sense to me that the inner diametre to be oversized and the rest of the dimensions don‘t matter as none of them matter if they‘re ‚wildly‘ off being they don‘t change or disrupt the rings function if they‘re wrong. So a lazy designer simply applied the same tolerances to everything.</p>
| 47325 | How to interpret tolerance in technical drawing |
2021-09-22T19:04:34.557 | <p><sub>Please do note that I am a civil engineer not a mechanical engineer.</sub></p>
<hr />
<p>I have a gear that rotates at a speed that varies with time. Usually, it rotates at speeds higher than the desired speed. Is there a mechanical mechanism that can regulate the output speed, i.e., a mechanism that can give a lower output speed without wasting power?</p>
<p>I am thinking about using a mechanism similar to the escape wheel in clocks but I am not sure if it can work with high torque and on a bigger scale.</p>
<hr />
<p><strong>Update:</strong> Thanks for all who contributed to answer my question. I have to add that the power that turning the gear (let's name it big gear) is coming from other small gears which gets there torque from twisted torsion springs. So I need the "big gear" to rotate in a constant speed so that all the torque coming from the small gears adds up. I hope the question is clearer this time.</p>
| |mechanical-engineering|gears|torque|mechanisms|springs| | <p>Before using a gear first of all you have to know <a href="https://www.mechical.com/2021/04/types-of-gears.html" rel="nofollow noreferrer">types of gears and their functions</a> then you can easily use without any obstructions.</p>
| 47330 | Mechanical mechanism that regulates gear's rotational velocity |
2021-09-23T15:18:57.123 | <p>I want to move a small tool table/bed linearly, so I'm designing its mechanism. Fairly simple, it's going to have 2 rails and a lead screw.</p>
<p>Browsing the web I find that locally the shortest trapezoidal thread rod I can get is 1m at the price of \$10 as well as \$8 for the nut and you can imagine that's me right now:</p>
<p><a href="https://i.stack.imgur.com/hdnqg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hdnqg.png" alt="enter image description here" /></a></p>
<p>So I was thinking, I have plenty of metric threaded rods and nuts laying around, and even if I had to buy new ones they cost next to nothing, but I've never seen one used as a lead screw for a mechanism that needs precise-ish movement. And looking at <a href="https://en.wikipedia.org/wiki/ISO_metric_screw_thread" rel="nofollow noreferrer">wikipedia</a> I found that metric threads are also trapezoidal, just at a steeper angle, so I'm guessing thats key.</p>
<p>Anyway, can a rod with metric thread be used as a lead screw with precision requirements of 0.5mm at worst?</p>
| |mechanisms|solid-mechanics|machining|threads| | <p>Well, the question has already been accepted, however. I'll mention something regardless.</p>
<p>First of all, it's perfectly fine to use a plain threaded rod for linear motion. You mostly simply need to consider what your basic requirements are, and your willing budget.</p>
<p>What I'm <em>most</em> surprised about, is that no-one here even considered to mention all the ways you can use ingenuity to solve the problem when money is a limiting factor. This was the common factor amoung ALL of the thousands of shitty 3d printer designs over the years, who all used threaded rod for various movements, and came up with tonnes of things to correct issues (like backlash). Some of them using software solutions or mechanical ones.</p>
<p>But first to list what you want:</p>
<ol>
<li>Linear movement, +/- 0.5mm precision</li>
<li>Cheap</li>
</ol>
<p>If you want to easily cheat, without spending any money at all, but are willing to get dirty, you simply go dumpster diving for old printers. They all have linear rails and different movement stages, some (most) have belts and pullys, some high end ones may have proper lead screws. Sometimes you can find really great parts from just old machines thrown away. There are countless backyard engineers who have made entire CNC machines using garbage they found.</p>
<p>This one of my favourite <a href="https://hackaday.com/2019/04/29/build-a-plotter-using-scrap-dvd-drives/" rel="nofollow noreferrer">examples</a></p>
<p>The next option is to consider which standards you're up against. Zinc coated threaded rod is really junk...But A4 stainless steel metric rod is <em>much</em> better. It's tolerances are much smaller and it's quality is generally very high while still being fairly cheap..</p>
<p>But you say, what about the backlash?</p>
<p>Well bachlash is a problem regardless of which screw system you use, even leadscrews can suffer from them (why do ballscrews exist?) So how have people over the years corrected for backlash at all?</p>
<p>Well there are tonnes of tricks for this.</p>
<p>Have a look at some mechanical libraries (never invent things twice if you don't have to, mech engineering has barely changed for hundreds of years, someone has probably already invented what you want to do)</p>
<p><a href="https://digital.library.cornell.edu/collections/kmoddl" rel="nofollow noreferrer">KMODDL</a></p>
<p><a href="https://www.dmg-lib.org/dmglib/main/portal.jsp" rel="nofollow noreferrer">Digital Mech Library</a></p>
<p>Here is a search result for <a href="https://www.dmg-lib.org/dmglib/main/portal.jsp?mainNaviState=search.results&page=1" rel="nofollow noreferrer">"backlash"</a></p>
<p>As quick tip, consider if you have 2 nuts on your rod that are tightened slightly in opposite directions on your rod....they can't have play if they are mechanically forced against that play can they?</p>
<p>Ofc there are problems with that...more friction, etc...but then we have oil right?</p>
<p>I also suggest becoming a regular reader of "<a href="https://hackaday.com" rel="nofollow noreferrer">hackaday.com</a>"</p>
<p>There are tonnes of great ideas on there.</p>
<p>Some search results there</p>
<p><a href="https://hackaday.com/blog/?s=threaded%20rod" rel="nofollow noreferrer">Threaded rod</a></p>
<p><a href="https://hackaday.com/blog/?s=backlash" rel="nofollow noreferrer">Backlash</a></p>
| 47353 | Does lead screw's thread have to be trapezoidal? |
2021-09-23T17:27:35.430 | <p>I'm working a closed system in which starts with 500 grams of steam and 500 grams of water at a temperature of 100 Celsius; 2199.26 kJ is put into the system. I'm trying to find the final temperature of the system.</p>
<p>Using <span class="math-container">$$Q(_1-_2) = E_2-E_1 $$</span>
I set <span class="math-container">$E_1$</span> equal to 0 since it's the initial state resulting in <span class="math-container">$E_2$</span> = <span class="math-container">$Q(_1-_2)$</span> which <span class="math-container">$E_2 = 2199.26 *kJ $</span></p>
<p>Using the assumption the rest of the water is turned to steam with the increased temperature, results in 1 kg of steam.</p>
<p>To convert the kinetic energy to temperature, dividing the energy by the mass
<span class="math-container">$$ \frac{2199.26* kJ}{1* kg} = \frac {2199* J}{1* g} $$</span></p>
<p>using the specific heat of steam <span class="math-container">$2.03 \frac{J}{gC}$</span>
results in
<span class="math-container">$$ \frac{2199* JgC}{2.03* Jg} = 1083.37*℃ $$</span></p>
<p>That seems a bit high while thinking logically and was wondering if I'm missing a step.</p>
| |thermodynamics| | <ol>
<li><p>We need the pressure of the mixture to know the latent heat of vaporization, but since we are using 100 C, we'll assume it's at 1 atmosphere.</p>
</li>
<li><p>Convert the liquid to steam using latent heat of vaporization. If your energy input isn't enough to turn it all into vapor, the final temp is 100 C. If you have energy left over it will go into the next step.</p>
</li>
<li><p>Using the specific heat capacity of steam vapor, calculate the temperature rise of the all-vapor steam from the remaining energy.</p>
</li>
</ol>
| 47359 | Converting kJ provided to a closed system into final temperature of the system |
2021-09-24T09:17:28.883 | <p>Today, as I was making -for another post - the following part</p>
<p><a href="https://i.stack.imgur.com/vOcZ5m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vOcZ5m.png" alt="enter image description here" /></a></p>
<p>Although, I managed to use a couple of ways to do the above (revolve a sketch, and b) using swept cut) I wondered how would I make grooves easily on a double curvature surface, e.g.</p>
<p>Something like :</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th></th>
<th>double curvature</th>
</tr>
</thead>
<tbody>
<tr>
<td>cymbal like</td>
<td><a href="https://i.stack.imgur.com/570hEm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/570hEm.png" alt="enter image description here" /></a></td>
</tr>
<tr>
<td>dome</td>
<td><a href="https://i.stack.imgur.com/B57f8m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B57f8m.png" alt="enter image description here" /></a></td>
</tr>
</tbody>
</table>
</div>
<p>(I don't know if there will be a difference, but the dome seems simpler somehow).</p>
<p>I would like to see how that would be done:</p>
<ul>
<li>along an axis of symmetry (Seems to me simpler)</li>
<li>along an offset</li>
</ul>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th></th>
<th>Axis of symmetry</th>
<th>Offset to axis of symmetry</th>
</tr>
</thead>
<tbody>
<tr>
<td></td>
<td><a href="https://i.stack.imgur.com/qfmK1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qfmK1.png" alt="enter image description here" /></a></td>
<td><a href="https://i.stack.imgur.com/hbjig.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hbjig.png" alt="enter image description here" /></a></td>
</tr>
</tbody>
</table>
</div> | |design|modeling|solidworks|cad| | <p>Placeholder answer - I've made the model, but don't have time to write up how I did it just now! :)</p>
<p>In short - make a plane, use intersection curve to make your sweep path. The, if you want to keep normal constant, make another path to use as a guide curve.</p>
<p><a href="https://i.stack.imgur.com/bdFG9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bdFG9.png" alt="Placeholder" /></a></p>
| 47376 | Solidworks - how to make a groove on a double curvature surface |
2021-09-24T13:28:06.650 | <p><a href="https://i.stack.imgur.com/IvNrg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IvNrg.jpg" alt="enter image description here" /></a></p>
<p>On the wall , i Walter had put a poster & when I removed the poster. I shone blue + red + white light on my white wall in the dark room.</p>
<p>Under blue light only , I am able to see so many details. Also , I see a reflection of a bit green color. It’s because of my camera it is looking like violet color. Not just wall but even on floor , the colour of dust appears to be green. Why is that ?</p>
<p>Under red & white light , I do not see these tiny details. Especially under red , I don’t see any of these details. Why is that ?</p>
<p>I do have studied about photo electric effect & think that it is somewhere used here. But I m not able to think how ?</p>
| |optics| | <p>As a guess, you might be picking up some fluorescence. Fluorescence absorbs light with a shorter wavelength and reemits is at a longer wavelength. Blue to green fluorescence is pretty common, but you are unlikely to notice it since the fluoresced color is so low compared to the excitation light. The room has to be dark and the excitation light needs to be a short wavelength where your eye and camera are less sensitive. Black lights which emit UV light make it easier to perceive fluorescence. Some white surfaces like high brightness printer papers have fluorescent brighteners. Fluorescence is a very important phenomenon in laboratory diagnostics testing.</p>
| 47384 | Why does my wall reflect green color when I shine blue light on it? |
2021-09-24T19:31:52.450 | <p>I'm in need of some gears, but having them made by a machinist is very expensive so I'm planning to have them laser cut. Loads are reasonably low for any grade of steel, so I'm hoping it should be fine.</p>
<p>Anyway, laser prices are given in amount per linear meter of cutting. In order to calculate how much it's going to cost me I need to calculate the perimeter of each gear and I just can not find how to do that...</p>
| |gears|geometry| | <p>Turns out there is no single formula for finding the circumference of a gear.</p>
<p>So I came up with one that approximates it with about 5% error which is more than enough for my purpose.</p>
<p><span class="math-container">$$P = 2\pi mz$$</span></p>
<p>Where <code>m</code> is module and <code>z</code> number of teeth.</p>
<p>How I came up with it: I drew a few gears and calculated their circumferences using the method described in <a href="https://engineering.stackexchange.com/a/47392/20333">Transistor's answer</a>. For each I calculated the pitch circle's circumference and observed how it relates to the already known real circumference - turned out the double of it matched all gears.</p>
| 47390 | How to calculate perimeter of spur gear? |
2021-09-26T09:33:42.133 | <p>I am designing a tractor implement for harvesting root crops. Similar to <a href="https://www.youtube.com/watch?v=OrFVcQOJlB0&ab_channel=BigElkGarlicFarm" rel="nofollow noreferrer">this one here</a>, but stronger construction with cutting discs in front of it to prevent weed jam. These images will help you understand my question. The whole implement is 1400mm wide and the pipe at angle is 200mm long relative to x axis.</p>
<p><a href="https://i.ibb.co/RYnxbRm/weld1.png" rel="nofollow noreferrer"><img src="https://i.ibb.co/RYnxbRm/weld1.png" alt="2" /></a></p>
<p><a href="https://ibb.co/FmvS3RM" rel="nofollow noreferrer">https://ibb.co/FmvS3RM</a></p>
<p><a href="https://i.ibb.co/QpdHRQZ/weld2-txtv2.png" rel="nofollow noreferrer"><img src="https://i.ibb.co/QpdHRQZ/weld2-txtv2.png" alt="3" /></a></p>
<p><a href="https://ibb.co/wBsQXgT" rel="nofollow noreferrer">https://ibb.co/wBsQXgT</a></p>
<p><a href="https://i.ibb.co/s1sJBLX/weld3.png" rel="nofollow noreferrer"><img src="https://i.ibb.co/s1sJBLX/weld3.png" alt="4" /></a></p>
<p><a href="https://ibb.co/SmV7LqZ" rel="nofollow noreferrer">https://ibb.co/SmV7LqZ</a></p>
<p>The main frame has 2 square pipes welded together at 45 degree angle at each side. The pipes are 120x120mm(4 3/4"x4 3/4") with wall thickness 8mm(0.315"). I was sceptical about using butt joint so I put a 15mm(5/8") steel plate between pipes. Now the pipes are welded with a fillet weld (at 90°, 45° and 135°). I'd like to hear some expert or practitioner opinion on how structurally stable welding pipes in such arrangement is with respect to flat pipe without any welds. The forces will act on the sides, that are just 20cm(8") from the three point hitch arms so the leverage is pretty small.</p>
<p>Thanks!</p>
| |structural-engineering|welding| | <p>You are complicating the matter and weaken your design by replacing the square pipe with a flat bent plate.</p>
<p>I don't get the reason you can't get the weld done if the weld angles conforming to the limits set in the figure below:</p>
<p><a href="https://i.stack.imgur.com/ZEOyH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZEOyH.png" alt="enter image description here" /></a></p>
<p>Note there are other ways to built-up the weld around this joint. Also, an angle smaller than <span class="math-container">$60^o$</span> is permitted; however, in such a case, the weld is considered to be a partial joint penetration groove weld.</p>
<p>AWS D1.1 Specification Section 2.11 Skewed T-Joints indicates:</p>
<p>"2.11.3.1 Z Loss Reduction. <strong>The acute side of prequalified skewed T-joints with dihedral angles less than 60° and greater than 30° may be used as shown in Figure 3.11, Detail D.</strong> The method of sizing the weld, effective throat “E” or leg “W” shall be specified on the drawing or specification. The “Z” loss dimension specified in Table 2.2 shall apply."</p>
<p>This article will help you in designing a welded T-joint. <a href="http://www.jflf.org/v/vspfiles/assets/pdf/design_file102.pdf#:%7E:text=not%20present%20%20%20%20%20%20,0.541%200.576%20%20%2014%20more%20rows%20" rel="nofollow noreferrer">http://www.jflf.org/v/vspfiles/assets/pdf/design_file102.pdf#:~:text=not%20present%20%20%20%20%20%20,0.541%200.576%20%20%2014%20more%20rows%20</a></p>
| 47418 | welding pipes with an angled fillet weld |
2021-09-26T12:14:54.363 | <p>I have seen uses of the word 'rectiplanar' in some older math books but Google-ing doesn't really give an answer to its meaning. I have found though the following <a href="https://pdfpiw.uspto.gov/.piw?PageNum=0&docid=08668756&IDKey=EBB97A352864&HomeUrl=http%3A%2F%2Fpatft.uspto.gov%2Fnetacgi%2Fnph-Parser%3FSect1%3DPTO2%2526Sect2%3DHITOFF%2526p%3D1%2526u%3D%25252Fnetahtml%25252FPTO%25252Fsearch-bool.html%2526r%3D1%2526f%3DG%2526l%3D50%2526co1%3DAND%2526d%3DPTXT%2526s1%3D20130199143%2526OS%3D20130199143%2526RS%3D20130199143" rel="nofollow noreferrer">patent application</a> where this word is mentioned and defined, but I am struggling to understand exactly what that means. The excerpt from the application is this:</p>
<blockquote>
<p>In figs 1-10, the panel filter element 38 is rectiplanar, i.e. lies in a single two-dimensional plane which is rectilinear in each of such two dimensions.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/bFyak.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bFyak.png" alt="enter image description here" /></a></p>
<p><strong>Figure : fig 3 from the patent application which shows element 38</strong></p>
<p>The link I have provided has these figures as well. This definition mentions that the element lies in a two-dimensional plane, but most of these pictures are 3-d. On figures 4 and 5 the panel filter element 38 looks to me like a 3-d object, so how can it lie in a 2-dimensional plane? Can anyone help me understand this definition?</p>
| |terminology| | <p><a href="https://i.stack.imgur.com/bgYTM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bgYTM.png" alt="enter image description here" /></a></p>
<p>Other than the noted shapes, the rest can be simply drawn on a 2D plane with a note t, L, or h = xx, thus "rectilinear element" or "rectiplanar element".</p>
<p>Everything on the earth is 3D, but depends on the complicacy of its geometry, very often a 3D element is identified as a 2D element, such as a beam, a plate...</p>
<p>Can you represent/describe the whole element in Fig 3 on a single (2D) plan?
The answer obviously is you can, but you are going to "note the details" which requires more effort than you simply draw more plans.</p>
| 47424 | What does the word rectiplanar mean? |
2021-09-26T14:01:53.457 | <p>A cylindrical container equipped with a manometer is inverted and pressed into water. The differential height of the manometer and the force needed to hold the container in place are to be determined</p>
<p><a href="https://i.stack.imgur.com/S5Bp5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S5Bp5.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/1l5LY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1l5LY.png" alt="enter image description here" /></a></p>
<p>Since pression in <span class="math-container">$A$</span> is the same as pression in <span class="math-container">$B$</span> i got</p>
<p>But since i cant find the value of <span class="math-container">$d$</span> i am unable to solve the rest of the exercice. Any suggestions?</p>
| |fluid-mechanics| | <p>the force need to hold the cylinder down is</p>
<p><span class="math-container">$$F=\pi \frac{25^2}{4} 20/1000cm*1.033kg/atm- \text{weight of the cylinder}$$</span></p>
<p>the difference between the two sides of the manometer is</p>
<p><span class="math-container">$$h=20cm/2.1=9.523cm$$</span></p>
<h1>Edit</h1>
<p>responding to OP's comment on how to measure the effective weight of partially submerged cylinder after deducting buoyancy.</p>
<p><span class="math-container">$$W_{submerged cylinder}= W_{dry} - \pi *25*20*t$$</span></p>
<ul>
<li>where t is the thicknesses of the cylinder. and weight is in grams.</li>
</ul>
| 47425 | The differential height of the manometer and the force needed to hold the container in place are to be determined |
2021-09-26T16:18:44.470 | <p>I have a 5 <span class="math-container">$m^3$</span> capacity tank that generates steam with a heat exchanger.
This tank is under 15 bar pressure. The water in the bottom part of the tank does not completely vaporize because there is always some fresh water filling it.</p>
<p>Under this bottom part, there is a purge valve connected to a 50 mm diameter pipe.</p>
<p>I would like to know how to estimate the water flow rate going though this purge pipe when the valve is open on a very short period of time (few seconds).</p>
<p>I tried to calculate the water velocity in the pipe with Bernoulli's equation but the numerical application gives me a really high value, which makes me think I made a mistake somewhere, or that this formula cannot be used for this installation:</p>
<p><span class="math-container">$V_{tank} = 0\,m/s\\
h_{tank} = 0,5\,m\\
P_{tank} = 1.500.000\,Pa\\
h_{pipe} = 0,5\,m\\
P_{pipe} = 0\,Pa\\
\rho = 866,89\,kg/m3$</span></p>
<p><span class="math-container">$\frac12 \rho V_{tank}^2 + \rho g h_{tank} + P_{tank} = \frac12 \rho V_{pipe}^2 + \rho g h_{pipe} + P_{pipe}$</span></p>
<p><span class="math-container">$V_{pipe}=\sqrt{\frac{2P_{tank}}\rho} = 58,3\,m/s$</span></p>
<p>And, I would also like to know how to find the water flow rate over time while estimating the pressure drop?</p>
<p>Any help would be much appreciated</p>
| |pressure|steam|bernoulli| | <p>If I am not seeing it wrong you have a tank that has a head of</p>
<p><span class="math-container">$$H_{vapor}= 15bars*10.2m/bar=153m$$</span></p>
<p><span class="math-container">$$Q=C_d A\sqrt2gh$$</span></p>
<p>where</p>
<ul>
<li><p>Q = flow (cubic metres per second)</p>
</li>
<li><p><span class="math-container">$C_{d} = coefficient\ of\ discharge$</span></p>
</li>
<li><p>A = area of orifice (square metres)</p>
</li>
<li><p>g = acceleration from gravity (9.81 m/s^2)</p>
</li>
<li><p>h = head acting on the centreline (m)</p>
</li>
<li><p><span class="math-container">$A = \pi/4(D^2)$</span></p>
</li>
</ul>
<p>Assuming the discharge opening doesn't have sharp edges, we pic Cd=0.8</p>
<p><span class="math-container">$$Q =\frac{0.8\pi}{4} (D^2)\sqrt{2gh}$$</span></p>
<p>After one second we have expanded the volume of 5m^3 by Qm^3. if we don't want to solve the dv/dt differential equation and ignore the friction of the 50mm pipe,</p>
<p><span class="math-container">$$P_1V_1=P_2V_2= P_2(V_1+Q)$$</span></p>
<p>After one second if we assume the expanded volume V2 is 5m^3 + Q which is a reasonable estimate,</p>
<p><span class="math-container">$$P_2=\frac{15bars*5m^3}{(5+Q)m^3}= 15bars \frac{5}{5+Q}m^3$$</span></p>
| 47430 | Water flow rate generated by a pressure steam tank |
2021-09-27T22:57:05.527 | <p>I'm trying to size a motor for an e-bike, but I'm confused as to what motor parameters determine the maximum torque that it can output (Kv or power in watts); I've found the following formulas:</p>
<p><strong>Power (watts) = torque (Nm) * angular_velocity (rad/s)</strong><br />
<em>also note that P = I(amps) * V(volts)</em></p>
<p>and</p>
<p><strong>Torque (Nm) = Kt * I (amps)</strong><br />
Where Kt is the torque constant or (1/Kv), and Kv is rpm/volt</p>
<p>So let's say we have a 1000 watt motor with a Kv of 200, the motor operates at 24V. Let's also say that we're talking about a low rpm (60 rpm);if we multiply rpm by (2 * pi / 60) then we get angular velocity. Using P = I*V we can calculate the max current to be 41.6 amps.</p>
<p>So according to the first formula, with a low rpm of 60 we could get a torque of <strong>~159 Nm</strong> at max power.</p>
<p>Using the second formula we get a torque of <strong>0.208 Nm</strong> at max power</p>
<p>These are massively different, and would require changes to gear ratios, battery sizes, material choices, etc.</p>
<p>So my question is: which one is right? Is torque determined by the Kv of the motor? or is torque determined by the power rating of the motor?</p>
| |electrical-engineering|motors| | <p>As you note yourself, Kv = 1/Kt. 200 Kv (RPM per volt) = ~21 rad/sec per volt. Invert this to get a torque constant of 0.04775 Nm per amp. This is what you want to know. As you calculated, at peak power, the motor will draw 41.7 amps. Multiply this by the torque constant to get about 2 Nm of torque.</p>
<p>The issue I see is that you are assuming that all of the motor ratings are valid across the entire curve. They are only valid at one point. We don't yet know at what RPM peak power occurs.</p>
<p>Let's use your example of a 1000 W, 24V, 200 Kv motor: 1000W is the peak power. But you don't yet know at what RPM this power is achieved. For an ideal motor it is when the input and output impedances are matched. At 24 V, and 200 Kv, the motor would spin at 4800 RPM no load (zero torque). Peak power would occur at 2400 RPM. But be careful. Motors can't always operate across the full range of torque and RPM. 24 V is the max allowable voltage. But there is also a max allowable current (41.7 A). So at RPM below 2400 you MUST limit the current to 41.7 Amps. You will not achieve more than 2 nM of torque. And total power will be reduced. At RPM greater than 2400, you must limit the voltage to 24 V. Current (and thus torque) will be reduced, and again, total power will be reduced. You can see that there is only one place where you can actually achieve peak power. See the example image below.</p>
<p><a href="https://i.stack.imgur.com/x5sa4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x5sa4.jpg" alt="Example motor torque and speed curve" /></a></p>
<p>Most motors are rated for operation at much higher speeds than peak power. Usually around peak efficiency. In this case the manufacturer should specify at what RPM peak power occurs.</p>
<p>Note that if you calculate power by your first equation using 2400 RPM (260 rad/s) and 2 nm, you get about 500 W. This is correct because the motor's power rating is the <em>input</em> power. When the input and output impedances are matched, half the power is dissipated in the motor, and half in the load (mechanical output). Sometimes motors are rated in terms of output power (usually for industrial or high performance uses), But then they MUST give you values for max allowable current as well as voltage. Reputable vendors will give you the exact torque curve and safe operating range.</p>
| 47450 | Is electric motor torque determined by Kv or by power (watts) |
2021-09-29T00:26:12.927 | <p>Let's say I have a box lying on a flat surface.</p>
<p>Inside the box, I make a heavy mass hang from a spring attached to the top of the box.
If we pull the mass all the way down and let it go, I believe it could have enough momentum to hit the top of the box and drag it up in the air a bit.</p>
<p><a href="https://i.stack.imgur.com/yjBmr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yjBmr.png" alt="enter image description here" /></a></p>
<p>I would like to make it more than once, so I'm thinking of putting a motor and worm gear or something, to be able to pull the spring down, lock it, then release, repeat (with a microcontroller)</p>
<p>I'm wondering about 3 things:</p>
<ol>
<li><p>If we don't pull the spring hard/far enough, when we release it it wouldn't provide much force. But if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right ?</p>
</li>
<li><p>What would be the formula to calculate the relationship between the movable mass's weight vs the whole box's weight, the spring's pulled length or tension and the net upward force ?</p>
</li>
<li><p>Is there a more efficient way than this approach, not requiring firing rockets ? :P</p>
</li>
</ol>
<p>Thanks !</p>
| |mechanical-engineering|structural-engineering|motors|dynamics|springs| | <p>1.) "... <em>if we stretch it too far, or keep it under tension for too long without firing, it would deform and then not work anymore, am I right?</em></p>
<p>In a general sense, your thoughts are fine, but let's be more specific about the different scenarios:</p>
<p>(1a.) - If the spring is stretched beyond its breaking strength, it breaks and would no longer work.</p>
<p>(1b.) - If the spring is stretched beyond its yield strength, it will have a permanent elongation in it that renders it become ineffective/defective.</p>
<p>(1c.) - If the spring is stretched beyond its ultimate tensile strength, it will continue to elongate without additional force until failure.</p>
<p>(1d.) - If the spring is stretched without the load removed for a long period of time, it will suffer the phenomenon called "creep" - an increase in strain without additional stress, and the time dependant strain is not recoverable (the spring will not return to its original length).</p>
<p>2.) "<strong>What would be the formula to calculate the relationship between the movable mass's weight vs the whole box's weight, the spring's pulled length or tension and the net upward force?</strong>"</p>
<p>I don't quite understand what is "relationship" you are looking for. Maybe the sketch below can offer some insight.</p>
<p><a href="https://i.stack.imgur.com/Pu5dW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Pu5dW.png" alt="enter image description here" /></a></p>
<p>In the sketch above, it is clear that in a place with the gravitational acceleration $g =0", as shown in the case (2a), the entire system (box + spring) will "float" in space if no external force is applied; if an external force is applied, the system will "drift" freely up or down, depends on the direction of the applied force.</p>
<p>3.) Is there a more efficient way than this approach, not requiring firing rockets?</p>
<p>Again, I don't quite clear on your approach. Do you mean to stretch the spring all the way down then release it? If so, the sketch below indicates it is possible <strong>if</strong> the rebound force (<span class="math-container">$k\Delta s$</span>) is greater than the total weight of the system. That is <span class="math-container">$\sum W < k\Delta s$</span>, then the system will bounce upward without external force.</p>
<p><a href="https://i.stack.imgur.com/mPql3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mPql3.png" alt="enter image description here" /></a></p>
| 47473 | How can you make a solid box jump up with no external forces? |
2021-09-29T00:54:24.143 | <p>I am testing the frequency response as a filter as below, what should be the sample rate of the chirp signals, someone told me it should be at least 2 times of the input signals frequency, someone said it should be at least 10 times? Why?
<a href="https://i.stack.imgur.com/qoQdA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qoQdA.png" alt="enter image description here" /></a></p>
| |control-theory|signal-processing|electronic-filters| | <p>2 times is the minimal sampling rate to ensure the signal can be possibly measured. This is called the Nyquist frequency. It shows that if your sampling rate is twice the frequency of the signal, you measure at least 2 different values per period. If you would only measure 1 value, a lower frequent signal or even a constant value could fit your measurement.</p>
<p>10, or even 20 times is what you should use if you desire to filter the signal (or for that fact, control it). This allows to fit your response so smooth, continuous-time filters can be properly applied without to many sampling issues.</p>
| 47474 | What should be the sample frequency to test the response of a filter? |
2021-09-29T19:03:59.373 | <p>Consider the spring mass dashpot system shown below which is acted upon by a harmonic excitation force F(t). The reference is taken at the equilibrium position of the system when no F(t) was present. At time t=0, F(t) acts and displaces the system from its equilibrium.</p>
<p><a href="https://i.stack.imgur.com/Ezkm2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ezkm2.jpg" alt="enter image description here" /></a></p>
<p>I'm interested in knowing the force transmitted to the support. I proceeded in the following manner: At time t, the spring would be stretched by <span class="math-container">$x-x_0$</span> and the end of damper will have a velocity equal to x(dot on the top)
<a href="https://i.stack.imgur.com/RzRUE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RzRUE.jpg" alt="enter image description here" /></a></p>
<p>The resulting expression I get for transmitted force has an mg in it. However all the sources I'm referring to state the expression with no mg. Where am I wrong in the analysis?</p>
| |mechanical-engineering|vibration|machine-elements| | <p>In the analysis you are referring to the authors are only concerned about the amplification of the excitation force (they are not concerned about the total force). I.e. they only care about the <strong>additional</strong> force that the excitation is causing on the supports.</p>
<p>(keep in mind that the vibration would still occur even in a zero-g environment if you added a harmonic excitation).</p>
<p>So, what they are doing is that they separate the static part (which is the mg) and the dynamic part.</p>
<p>It is easy to find the total force at the end by just adding mg to the total transmitted load (just like your equation).</p>
| 47485 | Force Transmitted to the support (Vibrations) |
2021-09-29T19:55:47.410 | <p>I have a friction damping system which is exited by a harmonic force FE (depicted on the left side).
Is there a way to convert the friction damper to a linear or nonlinear damper, such that the damping at a given excitement frequency is equal?</p>
<p>I am only considering sliding friction.</p>
<p>A reasonable approximation would be sufficient as well.
Any papers or articles on the topic would also be highly appreciated.</p>
<p><a href="https://i.stack.imgur.com/YkzOK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YkzOK.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|friction| | <p>In general this is impossible, because for a given value of <span class="math-container">$F_e$</span>, the friction damper dissipates a fixed amount of energy per cycle of vibration <em>independent of the vibration frequency.</em></p>
<p>This is (nonlinear) <em>hysteretic</em> damping, not (nonlinear) <em>viscous</em> damping.</p>
<p>The only way to approximate this with a viscous damper would be to make <span class="math-container">$C$</span> a function of both the <span class="math-container">$F_e$</span> and the frequency <span class="math-container">$\omega$</span>, which won't produce a useful equation of motion except in the special case where the machine only operates at one fixed frequency <span class="math-container">$\omega$</span>.</p>
<p>Aside from that issue, a general way to make the approximation is to model one cycle of the stick-slip motion of the friction damper and find the energy dissipated during the cycle. Then choose <span class="math-container">$C$</span> to dissipate the same amount of energy.</p>
<p>For a simple slip-stick damper you can do this from first principles, though the details are messy, and you need the complete equation of motion of the system - you haven't specified how the mass and/or stiffness are connected to the damper.</p>
<p>A more general approach is to use the so-called Harmonic Balance Method to produce a numerical approximation. There are many variations on the basic idea (and many research papers describing them!) but one implementation is the NLVib function in Matlab.</p>
| 47486 | Equivalent viscous damper for friction damping |
2021-09-30T16:17:16.890 | <p>Electric grids are volatile: As there's close to no storage capacity, power in must equal power out all the time.
One part of the management toolset is the various power sources attached to a grid. For the most part, it's relatively straight forward to find some "characteristics" governing these:</p>
<ul>
<li>Predictable cycles, like day/night for solar. (or tides, winter/summer, dry/wet seasons, etc.)</li>
<li>Irregular patterns with some degree of forecasts, like wind for wind power and clouds for solar.</li>
<li>Short therm buffering, like heat in thermal power plants.</li>
<li>Quick response ramping up and down, like hydroelectric power with magazines and diesel generators.</li>
</ul>
<p>What these have in common is that they are intuitive and governed by readily accessible physics. Nuclear power is not intuitive. While it's certainly possible to read up, it quickly becomes a journey down the rabbit hole of isotope burnoff, half-lives and an endless sea of details so deep there's little hope of understanding what the end result of it all is.</p>
<p>Nevertheless, I have tried to extract some general "characteristics" from what I have read:</p>
<ul>
<li><em>Some</em> nuclear power plants can do "load following", adjusting power production within minutes or even seconds. Those are apparently more costly to develop, and the majority of existing power plants are not capable of this.</li>
<li>Most reactors can ramp production up or down over single and double digit hour periods, adapting to for example peak daytime demands.</li>
<li>Nuclear powerplants are thermal power plants, so they can make use of the same short-thermal buffering of heat.</li>
<li>There are significant technological differences between nuclear power plants, so their characteristics may be varied.</li>
</ul>
<p>I am looking for characteristics at the most coarse level such as these (if they are even correct?)</p>
| |nuclear-engineering|electrical-grid| | <p>Re: load following, from <a href="https://en.m.wikipedia.org/wiki/Load_following_power_plant" rel="nofollow noreferrer">Wikipedia</a></p>
<p><em>Modern nuclear plants with light water reactors are designed to have maneuvering capabilities in the 30-100% range with 5%/minute slope. Nuclear power plants in France and in Germany operate in load-following mode and so participate in the primary and secondary frequency control. Some units follow a variable load program with one or two large power changes per day. Some designs allow for rapid changes of power level around rated power, a capability that is usable for frequency regulation.[6] A more efficient solution is to maintain the primary circuit at full power and to use the excess power for cogeneration.[7]</em></p>
<p><em>While most nuclear power plants in operation as of early 2000's were already designed with strong load following capabilities, they might have not been used as such for purely economic reasons: nuclear power generation is composed almost entirely of fixed and sunk costs so lowering the power output doesn't significantly reduce generating costs, so it was more effective to run them at full power most of the time.[8][9] In countries where the baseload was predominantly nuclear (e.g. France) the load-following mode became economical due to overall electricity demand fluctuating throughout the day.</em></p>
<hr />
<p>Others with this ability include AP1000 (US), CAP1400 (China), VVER1200 (Russia), APR1400 (Korea)</p>
| 47498 | What, roughly, are the characteristics of nuclear power in grid management? |
2021-10-01T19:42:07.770 | <p>How can I solve this beam. I was able to find the moments of each section, from there I need help. Please if someone could help me.
The method to use is Double Integration.
<a href="https://i.stack.imgur.com/2mbPr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2mbPr.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/qowsw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qowsw.jpg" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|structures| | <p>I am going to provide the hints for you to solve this problem.</p>
<p>Hint 1: Since the moment diagram has a singular point (change shapes) at the internal supports B & C, so you need to consider each segment (A-B, B-C & C-D) separately in writing the moment equations.</p>
<p>Hint 2: However, due to symmetry in both beam geometry and loadings, you only need to evaluate over 2 segments (A-B & B-midspan).</p>
<ul>
<li><p>For segment A-B, write the moment equation <span class="math-container">$M_1 = \dfrac{w_1x_1^2}{2}$</span>, for <span class="math-container">$x_1 = 0 - 3$</span>.</p>
</li>
<li><p>For segment B-midspan, <span class="math-container">$M_2 = M_1 - Rx_2 - \dfrac{w_2x_2^2}{2}$</span>, for <span class="math-container">$x_2 = 0 - 3.5$</span>.</p>
</li>
</ul>
<p><a href="https://i.stack.imgur.com/kDVy0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kDVy0.png" alt="enter image description here" /></a></p>
<p>The first integration of the moment (<span class="math-container">$EIv"$</span>) will yield <span class="math-container">$EIv'$</span> & a constant <span class="math-container">$C_1$</span>, and the second integration of the moment will yield <span class="math-container">$EIv$</span> & a constant <span class="math-container">$C_2$</span>. After successfully integrations, you can find the constant <span class="math-container">$C_1$</span> & <span class="math-container">$C_2$</span> by inserting the proper coordinate (<span class="math-container">$x$</span>) with the boundary conditions - <span class="math-container">$C_1$</span> represent rotation, which is "zero" at the support; <span class="math-container">$C_2$</span> represent deflection, which is "zero" at the support point as well.</p>
<p>Note, since you have two moment equations (<span class="math-container">$M_1$</span> & <span class="math-container">$M_2$</span>), you will get 4 constants. I suggest naming the constants as <span class="math-container">$C_{11}$</span>, <span class="math-container">$C_{12}$</span>, <span class="math-container">$C_{21}$</span> & <span class="math-container">$C_{22}$</span> to avoid confusion.</p>
<p>The procedure looks somewhat complicated and confusing, but once you start doing it, it will become easy and clear.</p>
| 47513 | Double Integration Method |
2021-10-03T18:14:00.720 | <p>I read about Saint-Venant principle which is about the effect of replacing forces by equivalent forces system that effect on the same small region of a rigid body, all examples I come across are on replacing axial force acting on a column by equivalent distributed force,so my question is about replacing distributed force on small region by point force in thin beams, is that considered an application of Saint-Venant principle?
<a href="https://i.stack.imgur.com/ZanbY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZanbY.jpg" alt="enter image description here" /></a></p>
<p>edited:
thanks to everyone tried to help me, I get many unsewrs in short time, may be because i have defficulties with English language those unswers were not clear to me, also i didn't explane my question well
finally i find this page which contain three exemples the second one (table exemple) is what i was asking about ... thank you again
<a href="https://www.fembestpractices.com/2020/11/saint-venants-principle-interpretations.html" rel="nofollow noreferrer">https://www.fembestpractices.com/2020/11/saint-venants-principle-interpretations.html</a></p>
| |statics|stresses|beam|mechanical|strength| | <p>This is an added effort in response to your question after your latest update.</p>
<p><a href="https://i.stack.imgur.com/PafHb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PafHb.png" alt="enter image description here" /></a></p>
<p>In the situation above, if a < d (beam depth), practically we can replace the uniform load with a concentrated load, and say it is justified by the Saint-Venant Principle.</p>
<p>However, it will run into problems with the actual behaviors from the beam theory if we proceed to check the beam internal forces, for example, the internal reactions at the mid-span:</p>
<p><strong>For the beam with the uniform load:</strong></p>
<p><span class="math-container">$M = \dfrac{wa}{2}*\dfrac{L}{2} - w*\dfrac{a}{2}*\dfrac{a}{4} = \dfrac{waL}{4} - \dfrac{wa^2}{4}$</span></p>
<p><span class="math-container">$V = \dfrac{wa}{2} - w*\dfrac{a}{2} = \dfrac{wa}{2} - \dfrac{wa}{2} = 0$</span></p>
<p><strong>For the beam with the concentrated load:</strong></p>
<p><span class="math-container">$M = \dfrac{wa}{2}*\dfrac{L}{2} = \dfrac{waL}{4}$</span>, and</p>
<p><span class="math-container">$V = \dfrac{wa}{2}$</span></p>
<p><strong>Conclusion:</strong> By comparing the results, we can see the danger of using the analogy of the Saint-Venant Principle, it can lead to erroneous results; except in very limited cases, or we can tolerate the mistakes from an approximation, or the calculations do not concern member internal forces/stresses, such as in the stability analysis, for which only the external forces/reactions are of interests.</p>
<p>However, Saint-Venant is not to blame for the limitation in its applicability, it's us to be blamed for not truly understand what engineering phenomenon it encompasses.</p>
<p>Hope this helps to erase your confusion over this matter.</p>
| 47540 | When we replace supports distributed reaction by point reaction in thin beams, is that considered an application of Saint-Venant principle? |
2021-10-04T04:12:38.590 | <p>I'm reading a root-locus course note from MIT. However, I can't understand this part:
<a href="https://i.stack.imgur.com/a6z09.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a6z09.png" alt="enter image description here" /></a></p>
<p>How does adding a compensator with a pole at origin remove the plant pole on the right half-plane into the left half-plane?</p>
| |control-engineering|control-theory| | <p>If I read it correctly (i.e. the plant has a single zero but not pole at the origin), then the issue I think is being solved, is that the plant RHP pole we are trying to manipulate is unpaired. Thus its RL would travel along the real axis. However the zero at the origin, being its destination at infinite gain, effectively blocks this pole's RL from ever reaching the LHP, as would be required for stability.</p>
<p>Adding an additional pole at the origin -- think of it instead as a RHP pole very close to the origin -- means you now have two RHP poles. So they will do the pole-splitting thing, where they converge on the positive real axis, then split up going up/down, and eventually get pulled into LHP (provided there are enough net zeros in the LHP to attract them).</p>
<p>The text actually describes direct cancellation, which is a more special case, in my understanding. Conceptually, for that I'd modify the above description to put the additional pole very slightly to the left of the origin. This results in traveling along the real axis again, and eventually a (very near) pole-zero cancellation in the CL. Whenever there is pole-zero cancellation, you have to sortof check the 'order of operations', because it has potential to conceal internal instability.</p>
<p>Also it's a little confusing, because the RL that is drawn only makes sense to me if there is a double zero at the origin.</p>
| 47548 | Could someone explain how a compensator shifts the root locus? |
2021-10-04T13:46:54.400 | <p>Is there a way to make fiberglass so safe that it can be used for everyday clothes?</p>
<p>From what I've found glass fibers are itchy when thick, so reducing their diameter helps make them soft on touch.</p>
<p>But In safety of asbestos and similar materials like carbon fiber I've found that reducing diameter below 6um makes them increasingly more carcinogenic, as such small shards are hard for the body to remove from the breathing system. 100nm are orders of magnitude more carcinogenic, danger when exposed for decades increases cancer rate tens of percents, and are similar in size to what asbestos has, but asbestos has a lot of short shards that are loose, so probably keeping all the threads long or impregnating them with rubber-like material that won't let shards loose might help? Are there attempts to make fiberglass clothes?</p>
<p>An example use where it could be very handy is tents.</p>
<p>Everyday clothes is a desired safety level, not an actual use case. I know that tent materials are often much less safe, and don't want this to be the case this time.</p>
| |threads|glass| | <p>Fiberglass is used in specialty clothing already. Cut resistant gloves for instance.</p>
<p>Also for thermal protection for welders, fire fighters, and lab rats.</p>
<p><a href="https://www.auburnmfg.com/product/ami-flex-afl-cloth/" rel="nofollow noreferrer">https://www.auburnmfg.com/product/ami-flex-afl-cloth/</a></p>
| 47550 | Safety of fiberglass, cloth |
2021-10-04T21:33:49.640 | <p><a href="https://i.stack.imgur.com/mXLlY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mXLlY.png" alt="angle image" /></a></p>
<p>The setup shows 3 rigid links (with fixed lengths) constrained by 4 joints. The blue joints are fixed in place and only allow rotation, while the black joints can move and also allow rotation. Initially θ1 < θ2, and the top link is perfectly horizontal.</p>
<p>As θ1 decreases (due to the leftmost link rotating clockwise), it is clear that θ2 will increase, but I would like to get the specific value of θ2 as a function of θ1. Is there a name for this kind of problem? What would be the best software tool to simulate this?</p>
| |mechanical-engineering|dynamics|kinematics| | <p>Since considering four-bar linkages is a useful tool in my research field, I happen to know of a paper that precisely solves what you want to do. George H. Martin published in 1958 in the journal <em>Machine Design</em> the paper "Four-Bar Linkages", the following equations are taken from that paper:</p>
<p>When you consider a four-bar linkage as shown below, then you know the angle <span class="math-container">$\theta _2$</span> and you want to know the angle that is the sum of <span class="math-container">$\beta$</span> and <span class="math-container">$\lambda$</span>.
<a href="https://i.stack.imgur.com/6KHXt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6KHXt.png" alt="Four-Bar Linkage" /></a>
Considering the triangle OAD, we can write
<span class="math-container">$$ \beta = \sin^{-1} \left( \frac{b}{l} \sin \theta_2 \right).$$</span>
So that is already the first half. For the triangle ABD, we can write
<span class="math-container">$$ \psi = \cos^{-1} \left( \frac{c^2+l^2-d^2}{2cl} \right).$$</span>
Using <span class="math-container">$\psi$</span>, we can write <span class="math-container">$\lambda$</span> as
<span class="math-container">$$ \lambda = \sin^{-1} \left( \frac{c}{d} \sin \psi \right).$$</span></p>
<p>With these, you should be able to work out a single equation for your problem.</p>
| 47556 | Is there a simple trigonometric relationship between these two angles? |
2021-10-05T04:28:21.213 | <p><em><strong>Context.</strong></em> Door size is about 1m wide and 2m tall, and environment is sunny, humid and hot climate.</p>
<p><em><strong>Question.</strong></em> What are the material options for the structure such that the door is durable? I will then study the options as per my budget and pick one.</p>
<hr />
<p><em><strong>My observation so far.</strong></em> One ancient option is to use exterior-durable wood species. But I think nowadays this option is becoming more of a luxury for achieving natural looks. I'm not sure if my observation is correct.</p>
<p>Other options include: using metals (which metals?) and uPVC. Not sure</p>
<p>As for the finish, I can finish them with any material. E.g. even if I go with steel structure, I can still nail/screw/glue wood-sheets on it to make it look like wood. Although I'm not sure how durable this option will be compared to a full solid wooden door.</p>
<hr />
<sub>
<p>This is a broader version of <a href="https://woodworking.stackexchange.com/questions/13136/how-to-pick-the-right-timber-for-external-doors">that</a> as <a href="https://woodworking.stackexchange.com/questions/13136/how-to-pick-the-right-timber-for-external-doors">that</a> is limited to wood species, while this exploring material options beyond wood.</p>
</sub>
| |materials| | <p>Exterior doors Have to be able to resist elements, be fire-rated (usually 20 minutes, but it could vary depending on type and location), and tolerate a high frequency of traffic and abuse of getting banged by large items, furnuture and appliances pushed through them.</p>
<p>Traditionally they were made solid-core from</p>
<ul>
<li>Alder.</li>
<li>Poplar.</li>
<li>Cedar.</li>
<li>Rustica Red Oak.</li>
<li>Cherry.</li>
<li>White Pine.</li>
<li>Hickory.</li>
<li>Mahogany</li>
</ul>
<p>Here in the US, the price of hardwood has gone up drastically, so many are using composite construction with a faux finish of look of some species.</p>
<p>Now there are doors made of hard particle boards and clad with sheets of metal or PVC.</p>
<p>we need to balance cost VS durability, functionality, and esthetics.</p>
| 47562 | Which materials to use to make exterior doors? |
2021-10-05T05:03:53.240 | <p>I am practicing with Inventor LT 2021 to get used to it, and learn more about how to manipulate things in it. I was practicing with the following sketch and I developed a bunch of questions about the sketch.</p>
<p><a href="https://i.stack.imgur.com/mD0FW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mD0FW.png" alt="Practice Sketch" /></a></p>
<p>I can see the thickness of the main plate is 11 (noted just to the left of the overall 65 dimension for depth).</p>
<ol>
<li><p><strong>How do I position this main plate in terms of depth from the
reference plane to the near face of the plate?</strong> I see a dimension
that looks like 51 but that makes no sense to me as it mean it would
be beyond the back face of the lower or left circles which are 38
(19+19 or 24+14) from the reference plane.</p>
</li>
<li><p>Is is safe to assume that the arm going from the middle to the upper circle has its back side flush with the front side of the main plate? Or is it notched/stepped in slightly? Is the back of the upper circle flush with the back of the arm?</p>
</li>
<li><p>The main plate is 11 thick, and the upper left circles are 14 thick. Assuming this was centered on the main plate that would 1.5 overhang each side. The general fillet note states R2 fillets. <strong>Is it possible to place an R2 fillet in that space?</strong> I was not sure if fillets had to be tangent to their mating surfaces.</p>
</li>
<li><p>At the back edge of the center, left and lower circles is a V mark.
I am assuming this has a specific meaning in the
machining/mechanical engineering world. <strong>What does the V mark mean?</strong> no
fillet? (I come from the structural/bridge engineering world so not
familiar with most symbols for parts)</p>
</li>
<li><p><strong>When drawing something that has a tolerance to it such as the left ream hole 14.27 - 14.32, what size should it actually be drawn at in the model?</strong> I have been picking a nice number that is within the range then navigating to the tolerance for it and entering the max value and the minimum value.</p>
</li>
</ol>
<p>I know if I was in an office I would go ask the engineer/designer/client for clarifications, but since this is just a practice sketch from an PDF online I don't have the source to ask.</p>
<p>This is what I have so far:</p>
<p><a href="https://i.stack.imgur.com/1k6er.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1k6er.png" alt="Status" /></a></p>
<p>For those wondering why I have not accepted <a href="https://engineering.stackexchange.com/questions/47205/how-to-place-a-sketch-plane-at-a-specific-geometric-location">my previous question</a>, I am planning to as soon as I work my way back to that question and try it again with some of the new techniques I have learned by stepping back by more than a few examples.</p>
| |technical-drawing|drafting|drawings|autodesk-inventor| | <p>I'm modelling this in SolidWorks, because it's what I have installed on this PC. Everything should be transferrable to Inventor, but ask if you have any questions.</p>
<p>Step 1 - create the central core. N.B. the '8mm' here is a driven dimension. The drawing shows 8, not 8.0, thus I'm inclined to thing the marked 8mm is a rounding error. Placing the hole concentric with the centre of the R9 makes much more sense, and would still give 8mm on the drawing for that dimension if set to 0DP.</p>
<p><a href="https://i.stack.imgur.com/6qhIC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6qhIC.png" alt="Step1" /></a></p>
<p>Step 2 - Create the Circles. I say create these next, since they are the functional geometry. The webs/joints between them are <em>driven by</em> these. If the design were to change, you want the webs to be late in the tree. Note the reamed diameter has an equation <code>=(9/16)*25.4)</code> to accommodate the inches problem. I have sketched on the front plane, and defined the 24mm offset starting position for the extrusion in the extrude feature. Having all the sketches on one plane will make the webs a bit easier to define, without as much 'projected geometry'.</p>
<p><a href="https://i.stack.imgur.com/FehSw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FehSw.png" alt="Step2" /></a></p>
<p>Step 3 -</p>
<p>Repeat Step 2 for the other two circles.</p>
<p><a href="https://i.stack.imgur.com/nyfIDm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nyfIDm.png" alt="Step3" /></a></p>
<p>Step 4 - Cut the 8mm hole in the bottom right cylinder. In order to avoid creating extra reference planes that aren't needed, I sketched this on the Right Plane, and defined the cut to start 'from vertex', picking up the centrepoint of the sketch which defined the cylinder, and then 'up to next'. This will dynamically rebuild to be cutting from the right place, to the right distance, regardless of how the model changes.</p>
<p><a href="https://i.stack.imgur.com/k7T13.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k7T13.png" alt="Step4" /></a></p>
<p>Step 5 - Extrude the 'arm'. This is defined by literally three dimensions, and the implied tangency and collinearity stuff that I mentioned in the other answer.</p>
<p><a href="https://i.stack.imgur.com/rBtn8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rBtn8.png" alt="Step5" /></a></p>
<p>Step 6 - Extrude the main plate. Note that the R51 is <em>not</em> concentric with the central core... This is defined as 11mm, starting from the back of the arm.</p>
<p><a href="https://i.stack.imgur.com/RcIpw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RcIpw.png" alt="Step6" /></a></p>
<p>Step 7 - Add fillets. I had to do this in a few steps in order to get the best looking fillets - order of operations matters. This is a whole 'nother topic/answer in and of itself, so I'm not going into it here...</p>
<p><a href="https://i.stack.imgur.com/KYl2J.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KYl2J.png" alt="Step7" /></a></p>
<p>Step 8 - Realize you've forgotten to cut the notches. Don't forget there is a mirrored copy just visible via hidden detail on the original drawing. This is that '51' dimension again...</p>
<p><a href="https://i.stack.imgur.com/8ESF6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8ESF6.png" alt="Step8" /></a></p>
<p>Step 9 - Admire your work.</p>
<p><a href="https://i.imgur.com/FcbeQBV.gif" rel="nofollow noreferrer"><img src="https://i.imgur.com/FcbeQBV.gif" alt="Step9" /></a></p>
| 47563 | Symbol and dimension question for practice sketch |
2021-10-05T12:54:14.617 | <p>Consider the arrangement shown below. It consists of a bolt, clevis and plate. Since, the clevis and the plate press against the bolt, the bolt will be acted upon by forces 1,2,3 (as shown).</p>
<p><a href="https://i.stack.imgur.com/70Tqz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/70Tqz.png" alt="enter image description here" /></a></p>
<p>There is something called as bearing stress which will act on the bolt in this case.</p>
<p>I don't exactly get what is this bearing stress. According to what I know about stress, when a body is acted upon by loads, we consider an area in the body and on this area internal resistive forces are developed to oppose external loads, then we calculate the intensity of these internal resistive forces on that area to determine the stress.</p>
<p><strong>The bearing stress in this bolt correspond to what internal resistive forces?</strong></p>
<p><strong>On which area these internal resistive forces are taken?</strong></p>
<p>I've thought a little about it, and came with this, but not quite sure if this is what is exactly meant by bearing stress. Does the bearing stress correspond to the internal resistive forces shown by red? at the midplane?</p>
<p><a href="https://i.stack.imgur.com/G6xLf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G6xLf.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-analysis|stresses| | <p>Bearing stress is essentially the contact pressure between two surfaces (the bolt and the joint in this example).</p>
<p>I find illuminating the following picture that shows the force flow.</p>
<p><a href="https://i.stack.imgur.com/VSz4X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VSz4X.png" alt="enter image description here" /></a></p>
<p><strong>figure (source <a href="https://www.chegg.com/homework-help/questions-and-answers/pin-clevis-joint-see-figure-33a-found-forces-act-pin-depicted-free-body-diagram-figure-p31-q4062409" rel="nofollow noreferrer">chegg</a>)</strong></p>
<p>how it is transferred though one part of the joint to the bolt and then to the other part of the joint.</p>
<blockquote>
<p>The bearing stress in this bolt correspond to what internal resistive forces?</p>
</blockquote>
<p>The bearing stress correspond the shear force diagram (Q). They are transverse forces that would cause bending.</p>
<blockquote>
<p>On which area these internal resistive forces are taken?</p>
</blockquote>
<p>different parts of the bolt experience difference load. for example above (if the system is static), the total force on (2) is equal to the force (1) + force (3). So the force flow (and the stress fields) distribution will look like the following image.</p>
<p><a href="https://i.stack.imgur.com/DLogE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DLogE.png" alt="enter image description here" /></a></p>
| 47570 | Bearing stress in Bolts |
2021-10-05T16:32:16.730 | <p>I'm currently studying oil extraction and learned that the engines are powered by electricity or diesel. Probably there are other types of fuel used as well. By engines I mean the pumps that extract the oil from under the sea. My question is: Is the oil platform able to use the oil they are pumping to power their own engines? I believe it would need to be able to "transform" oil into fuel on site and use it, is that possible?</p>
| |petroleum-engineering|fuel| | <p>Platforms are weight and space limited so there is generally no room for processing( and processing will require more crew and their facilities). Gas, oil and water are usually separated, often it is legal to dump the water, saves pipelining it to shore for processing. So a nat gas engine should be possible but I have not heard of them used on a platform. And it can require a complex of 2 or 3 platforms to separate gas, oil, and water. Especially in the US gulf, many platforms are unmanned and just send all production to shore. Some very productive deep water platforms will have a permanently anchored tanker which give more room for separating ,etc. They might separate out a diesel fraction, gasoline would be out of the question.</p>
| 47574 | Do sea oil platforms use the oil they extract to power their engines? |
2021-10-06T00:22:25.353 | <p>How does one find the zeta value or the frequency of an over damped response curve. I am used to determining things by using the log-decrement equation. Does that apply here even though there is no oscillations?</p>
| |mechanical-engineering|control-engineering|dynamics|vibration| | <p>Picking up from NMech's answer:</p>
<p>For the step response, integration constants can be expressed in terms of <span class="math-container">$\zeta$</span> and <span class="math-container">$\omega_0$</span></p>
<p><a href="https://i.stack.imgur.com/IrBnx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IrBnx.png" alt="foo1" /></a></p>
<p><em>According to <a href="https://lpsa.swarthmore.edu/Transient/TransInputs/TransStep.html#Case_1" rel="nofollow noreferrer">swarthmore.edu</a></em></p>
<p>Where <span class="math-container">$\omega_0 = \sqrt{\omega_1\omega_2}$</span> with <span class="math-container">$(\omega_1,\omega_2)$</span> being the two characteristic frequencies, and <span class="math-container">$K$</span> is the step amplitude</p>
<p>After that, I believe <span class="math-container">$\zeta$</span> can be extracted from any two points on the step response <span class="math-container">$(t_1,x_1)$</span> and <span class="math-container">$(t_2,x_2)$</span>, with the procedure made independent of <span class="math-container">$K$</span> and <span class="math-container">$\omega_0$</span> by looking at the ratio <span class="math-container">$x_1/K$</span>, <span class="math-container">$x_2/K$</span> and <span class="math-container">$t_2/t_1$</span> . For any (<span class="math-container">$x_1/K$</span>, <span class="math-container">$x_2/K$</span>), chosen arbitrarily, <span class="math-container">$t_2/t_1$</span> should correspond to <span class="math-container">$\zeta$</span>. A table could then be generated with a spreadsheet.</p>
| 47582 | Control System Response |
2021-10-06T07:13:43.307 | <p>I would like to look for a shaft collar like the one used in this photo of a cable robot:<br />
<a href="https://i.stack.imgur.com/oeSFH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oeSFH.jpg" alt="enter image description here" /></a></p>
<p>As you can see, there is a flexure that clamps the collar onto the shaft (as a shaft collar does). In addition, there are two screw holes where the collar can secure to a pulley.</p>
<p>When shaft collar is typed into google you get an infinity of these, so clearly I don't have good search terms or the correct name for this type of component: <a href="https://i.stack.imgur.com/bA9gl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bA9gl.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|machine-design| | <p>You can look for</p>
<ul>
<li><code>mountable shaft collar with flange</code>
or</li>
<li><code>flanged mount</code></li>
</ul>
| 47585 | What is a shaft collar called that is used to restrain a pulley? |
2021-10-06T12:48:00.817 | <p>G´Day,</p>
<p>Something got me thinking and I want to know if anyone else things this is possible.</p>
<p>I wonder if you can buy a electric air compressor, Connect it to your car/truck Battery, and connect the output into your engine to make it more powerful? You may also be able to put the lever inside the car so you can control when to give it that extra boost.</p>
<p>Does anyone else think this is possible since your adding extra air into the engine for combustion?</p>
| |mechanical-engineering| | <p>Yes it is quite possible</p>
<p>Electric Superchargers are a thing.</p>
<p>I think you will find that you will have trouble with always-on boost that you fight with the throttle valve, versus the typical crankshaft-powered blower. You will also find that you need a pretty large electric motor to power such a device. You will also find it hard to get air through the blower when it is off.</p>
| 47595 | Compressor + engine |
2021-10-07T18:04:29.003 | <p><a href="https://i.stack.imgur.com/iZPOG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iZPOG.png" alt="enter image description here" /></a></p>
<p>How do I calculate the total forcerequired for a linear actuator?
The image enclosed shows a series of door panels, each with a cam that i would like to connect to a linear actuator on the right hand side.</p>
<p>In order to spec the actuator I need to estimate the total force, how do i go about doing this?</p>
<p>There will be 7No panels in series.
<a href="https://i.stack.imgur.com/iZPOG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iZPOG.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-analysis|torque|linear-motion|linear-motors| | <p>Your sketch shows that the door panels are hinged balanced at the center.</p>
<p>This means they are at equilibrium in any position and can turn freely back and forth. And unless you need to flap them really fast, no angular momentum concerns.</p>
<p>All you need to deal with is the friction at the joints and links. And that depends on whether your system is put together well by correct bearings and connections or it's clunky and prone to jam.</p>
| 47617 | How do I calculate the total force required for a linear actuator? |
2021-10-07T20:10:37.607 | <p>I have spent way too much time in the last weeks/months implementing and writing my own finite element solver from scratch. I successfully implemented 2D and 3D elements and wanted to verify a few of my results. As an example, I went ahead and modelled a simple beam with two-dimensional quadratic quad elements.</p>
<p>Its dimensions are <span class="math-container">$500 \times 50 \times 1 [mm]$</span>. I loaded the beam on one end and constrained the other side. To be more exact, I constrained <strong>every</strong> node at the very left in both (x,y) directions.</p>
<hr />
<p>Due to <span class="math-container">$\nu=0.3$</span>, the von mises stresses are not meaningful for me. I am particularly interested in the analytical results of <span class="math-container">$\sigma_x$</span> and the simulation results.</p>
<hr />
<p>I did the computation by hand myself:</p>
<h2><span class="math-container">$$\sigma_{x,max} = \dfrac{M}{W} = \dfrac{F\cdot l}{\frac{I}{e_{max}}}= \dfrac{F\cdot l}{\frac{a^2b}{6}} = 1200 \text{Mpa}$$</span></h2>
<p>Looking at the simulation, it yields the following (displaying <span class="math-container">$|\sigma_x|)$</span>:</p>
<p><a href="https://i.stack.imgur.com/3awIl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3awIl.png" alt="enter image description here" /></a></p>
<p>If you look closer, you will see the highest values to be found at the boundary:</p>
<p><a href="https://i.stack.imgur.com/57QRU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/57QRU.png" alt="enter image description here" /></a></p>
<p>There seems to be a singularity at the top and bottom node which I constrained. The values next to those nodes have regular stresses around 1200 Mpa which is exactly what I computed.</p>
<p>My question is: How could I reduce this boundary node effect? Maybe my fem-code has some bugs and this is actually not supposed to happen. I am very happy for any kind of feedback.</p>
<h1>Edit 1:</h1>
<p>Every node on the left of the beam is constrained the following:</p>
<p><a href="https://i.stack.imgur.com/8igfp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8igfp.png" alt="enter image description here" /></a></p>
<p>The force on the right end of the beam is applied equally to all nodes (mid and corner-nodes). I know that this is not ideal but since i am mainly focusing on stresses at the boundary, I am fine with that.</p>
| |structural-analysis|stresses|finite-element-method| | <p>Whatever is causing this is a bug, either in your input, your your FE code, or your graphics display package.</p>
<p>Since we don't have access to any of those, we can't help you find it. The only thing we can do is tell you that it shouldn't happen!</p>
<p><strong>UPDATE</strong>: Reading the OP's comments, there seem to be three problems here.</p>
<ul>
<li>The constraints are not consistent with the intended stress and displacement field. Specifically, for a 2-dimensional model you only need to fix ONE degree of freedom in the Y direction (e.g. at the middle note of the beam). That will eliminate any issues caused by non-zero Poisson's ratio.</li>
<li>The loads are also inconsistent. For 4-node elements, the axial load on the two "corner" nodes at the right hand end should be half the load on each of the other nodes. For higher order elements, google for "equivalent nodal loads" to find how to calculate them correctly. In fact this mistake will have minimal effect on the OP's test problem, but that doesn't make it less wrong.</li>
<li>The algorithm you chose to calculate the element stresses is probably the worst possible one. For displacement-formulation elements it is a terrible idea to calculate surface stresses directly from the element shape functions.<br />
The standard way is to calculate the stresses at the internal points in the element where they are optimally accurate, and then extrapolate to the boundaries. For a 4-node element, calculate the stress at the element centroid and assume it is constant over the whole element.<br />
For higher order elements, the optimal points depend on the integration rule used. The optimal points are sometimes referred to as "Barlow points" after their discover (see <a href="https://onlinelibrary.wiley.com/doi/abs/10.1002/nme.1620100202" rel="nofollow noreferrer">https://onlinelibrary.wiley.com/doi/abs/10.1002/nme.1620100202</a> for the original paper).</li>
</ul>
<p>To summarize all this: inventing a subject as complex as Finite Element Analysis all by yourself is a fun project, but if you want to produce <em>useful</em> software, you need to learn to avoid the pitfalls that others have discovered during the last 50 years or so.</p>
| 47621 | FEM boundary problem in structural mechanics |
2021-10-08T19:45:44.930 | <p>I work as a Design Engineer for an industrial ventilation company. In order to complete our outlet for our ventilaion system, we need to install a 54" by 40" steel panel (16 gauge mild steel) onto the roof (think of a skylight on a ceiling for example). My question is: could a 400lb man safely stand atop this panel without falling through? What would be the proper sizing of bolts installed around the perimeter of the panel to accomplish this? Thank you!</p>
| |mechanical-engineering|structural-engineering| | <p>If this is a roof hatch, it shall have its own frame to support the steel panel. You shouldn't have the problem figuring out the required bolts and bolt size by modeling a flexible beam between the hinge supports. The important thing is not only knowing the person will not fall through, but the deflection of the plate also counts heavily on human perception of safety.</p>
<p>Are you a drafting/design technician? If so, this task (bolting and framing) is the engineer in charge's, or your supervisors', job.</p>
<p><strong>Typical Roof Opening/Hatch Support Detail:</strong></p>
<p><a href="https://i.stack.imgur.com/aa4R2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aa4R2.png" alt="enter image description here" /></a></p>
| 47631 | Sheet Metal Panel Install on Roof |
2021-10-08T21:44:55.290 | <p>I am wondering how much longer the RMS Titanic could have remained afloat if the crew had allowed the ship's anchor and anchor chain to fall to the bottom of the ocean immediately after the ship had hit the iceberg. (I am not even sure if a ship's anchor chain can be unfastened from a ship, but let's just say for the sake of this question that it can be unfastened.)</p>
<p>The combined weight of Titanic's anchor and anchor chain was approximately 116 tons according to this Wikipedia article:</p>
<p>"...In 1911, the company manufactured the anchors and chain for the ocean liner RMS Titanic. The largest of the anchors weighed 15.5 tons and on completion was drawn through the streets of Netherton on a wagon drawn by 20 shire horses.[15] The chain and fittings for the anchors weighed around 100 tons..." <a href="https://en.wikipedia.org/wiki/N._Hingley_%26_Sons_Ltd" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/N._Hingley_%26_Sons_Ltd</a></p>
<p>Since the Titanic went down bow first, and the anchor and chain was located in the bow section, immediately getting rid of 116 tons in the bow section would have increased the time it had remained afloat before it sank.</p>
| |mechanical-engineering|fluid-mechanics|marine-engineering|ships| | <p>The ship began to flood immediately, with water pouring in at an estimated rate of 7 long tons (7.1 t) per second. This means 116/7.1 = 16.33 second delay in sinking. But of course releasing the anchor also reduces the water pressure, so it can be more.</p>
| 47635 | How much longer could Titanic have remained afloat if it had gotten rid of its anchor and chain right after hitting the iceberg? |
2021-10-08T23:15:11.720 | <p>I'm looking for the common name(s) or other unique information of a 2 DoF pendulum joint that has one rod that pivots about one axis, connected to a fork that pivots about the orthogonal axis, which mounts to a fixed frame. The main rod extends all the way through the joint. Moving the rod in the x/y direction above the joint gives inverted movement on the opposite side.</p>
<p>Below is the best example of such a joint I could find. Here the rod appears to be in two parts (threaded rod on top, black plastic below), but they move as one piece.</p>
<p><a href="https://i.stack.imgur.com/6z4Wv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6z4Wv.png" alt="Joint" /></a></p>
| |mechanical-engineering| | <p>These joints are called gimbals. The main idea is that the pivots are nested. There is an outer pivot, and the frame inside contains another pivot, and so on.</p>
<blockquote>
<p>A gimbal is a pivoted support that permits rotation of an object about
an axis. A set of three gimbals, one mounted on the other with
orthogonal pivot axes, may be used to allow an object mounted on the
innermost gimbal to remain independent of the rotation of its support
(e.g. vertical in the first animation). For example, on a ship, the
gyroscopes, shipboard compasses, stoves, and even drink holders
typically use gimbals to keep them upright with respect to the horizon
despite the ship's pitching and rolling.<br />
<a href="https://en.wikipedia.org/wiki/Gimbal" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Gimbal</a></p>
</blockquote>
| 47637 | What is this 2 DoF joint called? |
2021-10-10T10:40:30.777 | <p>Imagine a block of mass <span class="math-container">$m$</span> placed on the ground. I slowly lift it up to height <span class="math-container">$h$</span>. The work done by me on the block is <span class="math-container">$mgh$</span>. This amount of energy is now stored in the block when it is at height <span class="math-container">$h$</span>. This energy is called the potential energy of the block. From where did we get the word <em>potential</em>? Why is it called <strong>Potential</strong> energy?</p>
| |mechanical-engineering|applied-mechanics|energy| | <p>In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, attachments, electrical charge, magnetic field, mass, etc., and external forces, spring force, electric charge, magnetic field, gravity, or others.</p>
<p>The term potential energy was introduced by the 19th-century Scottish engineer and physicist William Rankine, although it has links to Greek philosopher Aristotle's concept of potentiality. The concept of potentiality, generally refers to any "possibility" that a thing can be said to have. Aristotle did not consider all possibilities the same, and emphasized the importance of those that become real of their own accord when conditions are right and nothing stops them. Actuality, in contrast to potentiality, is the motion, change, or activity that represents an exercise or fulfillment of a possibility, when a possibility becomes real in the fullest sense.</p>
<p>The work of potential forces acting on a body that moves from a start to an end position is determined only by these two positions and does not depend on the trajectory of the body. The potential field can be evaluated at those two positions to determine work. This allows the set of forces to be considered as having a specified vector at every point in space forming what is known as a vector field of forces, or a force field. Furthermore, the force field is defined by this potential function, which is also called potential energy.</p>
<p>The term "potential" itself is a historical term that was carried into current practice, through the normal accident and convenience of the development of the physics and mathematics, from the natural philosophy at its root.</p>
<p><a href="http://web.ecs.baylor.edu/faculty/lee/ELC5360/Lecture%20note/Potential%20Energy.pdf" rel="noreferrer">http://web.ecs.baylor.edu/faculty/lee/ELC5360/Lecture%20note/Potential%20Energy.pdf</a>
<a href="https://en.wikipedia.org/wiki/Potential_energy" rel="noreferrer">https://en.wikipedia.org/wiki/Potential_energy</a></p>
| 47649 | Why does "potential energy" have the word "potential" in it? |
2021-10-09T18:59:28.153 | <p>Here is a simple schematic of the engine.<br />
<a href="https://i.stack.imgur.com/4uIIb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4uIIb.png" alt="enter image description here" /></a></p>
<p>The increase in angular momentum of the crankshaft in one complete cycle would be</p>
<p><span class="math-container">$\Delta L_{cycle}=\int_{\delta =0}^{\delta =4 \pi} F cos \theta Rdt$</span>.</p>
<p>where F is positive in combustion stroke and negative in other strokes (as L increases only in the combustion stroke and decreases in other strokes where work is done by the piston to draw in air-fuel mixture, compress the mixture, or expel exhaust gas). Also note, F would be different for each value of from 0 to 4.</p>
<p>The torque generated by the engine would then be
<span class="math-container">$$\tau = \frac{\Delta L_{cycle}}{\Delta t_{cycle}}$$</span></p>
<p>Now, let's see what happens when the angular velocity (ω) of the crankshaft is doubled.<br />
Any force F specific for an angle would then act on the crankshaft for only half the period of time (i.e., dt for each angle gets halved). As a result, the increment in angular momentum of the crankshaft in one complete cycle gets halved. However, since the time period of each cycle has halved as well, the torque generated by the engine should theoretically be the same regardless of the rpm of the engine. This would result in the following torque-rpm graph.</p>
<p><a href="https://i.stack.imgur.com/r1Sc3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r1Sc3.png" alt="enter image description here" /></a></p>
<p>However, we know that this is not the case. The actual graph looks something like this.
<a href="https://i.stack.imgur.com/mwGQo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mwGQo.png" alt="enter image description here" /></a>
The dip in torque is easy to explain as with increasing rpm one can expect several losses to increase such as frictional losses, pumping losses, inefficient charge mixing, decreased time for efficient combustion etc. But my question is, what explains the increase in torque with angular velocity at the low rpm range?</p>
| |thermodynamics| | <p>In addition to the volumetric efficiency of the valve timing system, there are some harsh losses associated with thermal losses. The top of the cylinder is hotter than the bottom end, and heat transfers down the cylinder liner at a rate that doesn't depend on rpm. So more heat is wasted per stroke (which is a poor way of saying it) when the engine is running slowly.</p>
| 47670 | Why does the torque of an internal combustion engine increase with its rpm at low rpm range? |
2021-10-11T18:25:10.440 | <p>I want to do a test where I use composite materials to mimic impact damage on aircraft materials such as impact from hailstone while the aircraft is in flight.</p>
<p>I will cut out specimen sheets using a water jet cutter and they will have different size notches and then I will use a drop weight impact machine to create a hole in the material and assess any cracks or deformations etc.</p>
<p>Here is the design of the specimen sheets that I will cut out from the larger material sheet:</p>
<p><a href="https://i.stack.imgur.com/TqwRm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TqwRm.jpg" alt="enter image description here" /></a></p>
<p>My question is what are some good objectives that I can set for this experiment before the test is carried out.</p>
| |mechanical-engineering|materials| | <p>Don't forget to maintain the temperature while carrying out the test. The material will behave differently when subjected to impacts on different temperatures.</p>
<p>Also, if the part you are experimenting on is the outside surface in contact with the air, then also don't forget to put on the necessary paint or any other surface coating that is originally used on it. This can also create a difference.</p>
<p>After cutting out the necessary specimens, make sure that the cuts are done gently and the cutted faces are smoothened enough to mimic the real life structure. The reason being after the cut out, the amount of pre-present cracks in the structure can increase and may make it more vulnerable to fracture than what can be expected in reality.</p>
<p>Since you are trying to model hail by dropping weights, you have to ensure that the weights has somewhat the same Elastic Modulus and rigidity as the hails. The reason being, if the weights are more rigid, then the cracks or even fracture can appear on the composite sample while it may not the case when hail of same weight is dropped on it. (Since in the situation of impact where drop weight is more rigid than a hail, all of the energy during impact will be taken by the composite sample).</p>
| 47676 | Impact damage in composite materials |
2021-10-13T12:41:08.193 | <p>In the book <strong>Fluid Mechanics by Yunus Cengel</strong> it was written</p>
<p><strong>"Flow around an airplane, for example, is always unsteady with
respect to the ground, but it is steady with respect to a frame of reference
moving with the airplane at cruise conditions."</strong></p>
<p>So my question is that why does this happens? And does this has anything to do with moving and stationary control volume?</p>
| |fluid-mechanics|aerodynamics| | <p>With respect to the ground, an airplane tears up the air it goes through - the disturbance isn't uniform it's unsteady, it moves across the world.</p>
<p>With respect to the airplane, the air flows around it the same way all the time. Airflow around the plane is uniform and steady.</p>
| 47694 | Why flow around an airplane is different in different frames of reference |
2021-10-14T15:38:21.847 | <p>I would really like to buy this textbook, but only see a scan of one chapter online. Could anybody help me identify the book? Thanks!</p>
<p><a href="https://i.stack.imgur.com/XccOa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XccOa.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|heat-transfer|hvac|refrigeration|dehumidification| | <p><a href="https://www.pearson.com/us/higher-education/program/Kuehn-Thermal-Environmental-Engineering-3rd-Edition/PGM221675.html?tab=contents" rel="nofollow noreferrer">Thermal Environmental Engineering</a> 3rd Edition by Kuehn, Ramsey & Threlkeld (1998), Pearson.</p>
<p><a href="https://i.stack.imgur.com/O7uDB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/O7uDB.png" alt="enter image description here" /></a></p>
| 47709 | Identify this textbook |
2021-10-14T20:25:50.673 | <p>I have the following question</p>
<p><a href="https://i.stack.imgur.com/QuRAs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QuRAs.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Ut7E3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ut7E3.jpg" alt="enter image description here" /></a></p>
<p>I used abaqus software to create a sketch and added the downwards load on member BC.</p>
<p>However the value I get for the stress is 90185.7</p>
<p>Is my answer incorrect because it is not in pascals or is my answer just completely wrong?</p>
| |abaqus| | <p>Your answer seems correct (in magnitude), verified by a quick check on the stress of member AB shown below. You shall check/confirm the stress in member BC by hand calc though.</p>
<p><a href="https://i.stack.imgur.com/Lczz1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lczz1.png" alt="enter image description here" /></a></p>
| 47714 | Calculating stress in a member |
2021-10-15T09:01:28.280 | <p>I have built <a href="https://www.youtube.com/watch?v=3VfiC85VjXM" rel="nofollow noreferrer">purely mechanical pinball machines</a> before which used score compartments the balls would fall into to keep track of points. I am now trying to find a way to count points without losing the balls, i. e. a mechanical counter that is triggered by the ball.</p>
<p>I should add that I am using simple tools (wood, bandsaw) and want to avoid ready-made parts as much as possible. Also, this is a mini-pinball machine (69 cm x 34 cm), so everything has to be rather small.</p>
<p>I was thinking of building a ratcheted wheel which would be under rotational force (I'm thinking weights as springs don't provide a constant force) and held by a spring-loaded arm. The ball passing that arm should temporarily release the ratchet to advance the wheel by exactly one notch. I have pictures attached of my prototype. My problem is that I need exactly the right balance of forces in order for it to work, which is super fiddly and I probably won't get it to work reliably. Does someone have an idea for a reliable "one-notch-advance"? Will I essentially need to build a <a href="https://en.wikipedia.org/wiki/Escapement" rel="nofollow noreferrer">clock escapement</a>?
<a href="https://i.stack.imgur.com/s2dio.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/77Btk.jpg" alt="Prototype overview" /></a>
<a href="https://i.stack.imgur.com/s2dio.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s2dio.jpg" alt="Prototype detail" /></a></p>
| |applied-mechanics| | <p>I agree with going for an escapement mech. However, your comment about not wanting to use a spring seems odd, seeing as springs are what power clocks & wristwatches, not to mention a lot of resettable counting wheels like this in electromechanical commercial pinball machines.</p>
<p>So, maybe if installing an escapement doesn't help, consider an "amplifier" analog. That is, the ball rolls against a lever with minimal resistance, which then triggers a secondary mech via level-length or gearing power gain.</p>
| 47718 | mechanical counter using ratchet (for a pinball machine) |
2021-10-15T16:26:19.683 | <p>I'm trying to design a powered swing for adults (similar to a baby swing, but scaled up). Am I right in assuming that this is functionally a pendulum and the power required is (0.5)mgsin(t)? How should I go about finding an appropriate motor?
I'm sorry if this sounds basic, but this is the first project I've ever done from scratch, so any help would be appreciated.</p>
| |design|motors|machine-design| | <p>Just as a fast estimate we assume the following:</p>
<ul>
<li><p>the only friction you swing experiences is air friction and l=2m, g=10ms^2.</p>
</li>
<li><p>that <span class="math-container">$ \ \omega =\sqrt{\frac{g}{l}}=\sqrt5=2.24rad.s^-1$</span></p>
</li>
<li><p>we assume you set the swing to travel an angle of 2.24/2rad=1'12rad, so its frequency of 1 second. and it travels with an average speed of 2.24*2m=4.48m/s (not correct but reasonable)</p>
</li>
<li><p>that the surface area of an adult sitting on the swing is 1 m^2</p>
</li>
</ul>
<p>Now we calculate how much force we need to keep the swing moving against the air friction.</p>
<p>Drag force is <span class="math-container">$F=\frac{C_d*A*\rho*v^2}{2}$</span></p>
<ul>
<li>Assume C_d =0.75 for a sitting person</li>
<li>Air density 1.225kgm^3</li>
</ul>
<p><span class="math-container">$$F=\frac{0.75*1*1.225*4.48^2}{2}=9.2N\ per\ second$$</span>
<span class="math-container">$F*1.5_{safety factor}= 13.8watt.$</span></p>
<p>Just a rough estimate.</p>
| 47723 | Motor For powered adult swing |
2021-10-15T21:44:14.690 | <p>Continuing through my practice examples I came across the following diagram and I have a question on how to interpret the depth of the part.</p>
<p><a href="https://i.stack.imgur.com/s5O5o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s5O5o.png" alt="Exmaple 103" /></a></p>
<p>I have the part drawn up in inventor, but when I went to do my extrusion I started interpreting the spacing between the plates in different ways. What is throwing me for a loop is the note "BAR DIA.70 x 76 LONG" and the 28 dimension.</p>
<p>I have interpreted it these ways:</p>
<ul>
<li>9.5 sticking out the front.</li>
<li>19 thick front plate.</li>
<li>12 thick rear plate.</li>
</ul>
<p><strong>Option A)</strong></p>
<p>From the face of the 9.5 protruding 66 dia. cylinder (assumed machined down to) to the back of the cylinder is 76 mm and there is a 28 mm space between the plates.</p>
<p><strong>Option B)</strong></p>
<p>There is a 76 mm cylinder between the plates and it protrudes 28 mm beyond the back of the back plate.</p>
<p><strong>Option C)</strong></p>
<p>There is a 76 mm cylinder starting at the back of the front plate that is 76 mm long and a 28 mm space between the plates.</p>
<p><strong>What is your interpretation for the setting the various depth?</strong></p>
<p>Option D) being something I have not considered and please provide details.</p>
| |technical-drawing|drafting|drawings| | <p><strong>There is not enough information in the drawing to give a definitive answer</strong></p>
<p>The question effectively is, which of the three highlighted surfaces should serve as the "start point" for the "76 long" dimension, given that we can't see the back side of the part.</p>
<p><a href="https://i.stack.imgur.com/bYAlj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bYAlj.png" alt="Options" /></a></p>
<p>In my <strong>opinion</strong>, it should be the very front face of the part.
My reasoning for this, is that the note here is intended for the fabricator, to allow them to select the correct stock material for the job. The inner surface is smooth without any join, so the bore is made of a single piece of stock material. The 66mm diameter is simply a segment of the 70mm diameter stock material, which has been machined down to fit.</p>
<p><a href="https://i.stack.imgur.com/OwuSd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OwuSd.png" alt="Smooth inner face" /></a></p>
<p><a href="https://i.stack.imgur.com/UQH5o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UQH5o.png" alt="My interpretation" /></a></p>
<p><a href="https://i.imgur.com/C8kpQlU.gif" rel="nofollow noreferrer"><img src="https://i.imgur.com/C8kpQlU.gif" alt="Animation" /></a></p>
| 47731 | Interpretation of dimensions |
2021-10-16T01:51:50.447 | <p>I remember hearing awhile back that it's not a good idea to use screws to locate/align parts. Why is that? Apparently dowel pins, or "locating pins" should be used instead. From my understanding, I suppose it is because there are much tighter tolerances when manufacturing those pins compared to fasteners.</p>
<p>Does this inherently mean that you should not use dowel/locating pins in a design when you are screwing into a threaded hole? Since there might be binding. In addition, if you were to decide you wanted to use locating pins/dowel pins for a more precise connection, then your holes for your fasteners should be not threaded (thru holes) and sized a bit larger than the shank diameter to allow some movement.</p>
| |fasteners| | <p>Our policy for assembling modest precision electronic instrument chassis was to have one countersunk screw that would locate two parts, and any other screws retaining them together would have a head that was flat underneath, to allow a tolerance in hole centres to accommodated.</p>
| 47736 | Why can't screws be used for locating parts? |
2021-10-16T03:04:52.437 | <p>We can consider that a ship is a complex implementation of forces on a beam, that a ship is essentially a beam.</p>
<p>The support is the buoyancy force which is continuous but it takes different values depending on the geometry of the ship, on the surface that sees the water.</p>
<p>The loads also differ along the length of the ship.</p>
<p>Could we name this a statically indeterminate structure?</p>
| |mechanical-engineering|structures| | <p>A ship is a statically indeterminate structure. Because:</p>
<blockquote>
<p>the static equilibrium equations – force and moment equilibrium conditions – are insufficient for determining the internal forces and reactions on that structure.</p>
</blockquote>
<p>Quote from Wikipedia. <a href="https://www.wikiwand.com/en/Statically_indeterminate" rel="nofollow noreferrer">source</a></p>
<p>One rule of thumb is when the support reactions and member forces are depending on the structure configuration, or its member section properties, the compatibility in deformations, that structure is indeterminate. A ship is an example of this.</p>
<p>A simple three-member triangular truss becomes indeterminate as soon as a vertical redundant member is added from the top vertex to the middle of the bottom member.</p>
<p>A ship by design has many structural redundant members. Thus it is an indeterminate structure.</p>
<p>An example of an indeterminate beam. It has 4 unknowns, but only three equilibrium equations. Details in the article.</p>
<p><a href="https://i.stack.imgur.com/0jexV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0jexV.png" alt="indeterminate beam." /></a></p>
| 47738 | Is a ship a statically indeterminate structure? |
2021-10-16T04:00:52.477 | <p>I am building an evacuated tube array for a solar water heater and I am selecting the materials to use for the tubing and the collectors (the places connected to the tubing that help absorb the solar radiation).</p>
<p><a href="https://i.stack.imgur.com/S3CU7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S3CU7.png" alt="terrible schematic" /></a>
I can choose from copper, aluminum, or steel as they are all relatively cheap and easy to work with the tools that I have on hand.
Everything else being equal, which of those three materials is best suited for collecting and transferring the heat to my 40% propylene glycol solution?
Additionally, can I add a coating, such as black paint, to increase the effectiveness of this material?</p>
| |thermodynamics|heat-transfer|thermal-conduction|solar-energy|solar| | <p>Copper sheet with an absorbing and poor emitting coating is available which is easy to solder copper pipe to the untreated reverse side. Guess why many flat plate collectors are made like this…</p>
<p>As for spacing you can get full details from Duffie & Beckmann Solar Thermal Engineering (can’t remember exact title, but it is a really good book) but panels I have made have about a 20cm spacing on a 1 or 1.2mm sheet which is about 90cm wide.</p>
| 47740 | Best metal for solar energy collection in an evacuated tube solar water heater |
2021-10-16T12:52:04.473 | <p><em><strong>Question.</strong></em> What non-welding options do I have to reliably attach square aluminium tubes to an aluminium frame?</p>
<p><em><strong>Context.</strong></em> This will be used for outdoor environment (gate). I'm thinking of non-welding options to attach horizontal square aluminium tubes to an aluminium frame.</p>
<p>The frame is welded. But I'd like the tubes to not be welded.</p>
<p>My reason is that I think that replacing non-welded aluminium square tubes is easier than re-welding. Because, repairing welds will require to repaint the door, including other areas near the weld, even parts not relating to the damaged part. On the other hand, if the square tubes are attached in a non-welding way, I can only replace the damaged square tubes, without needing to repaint anything nearby.</p>
<hr />
<sub>
<p><em><strong>Appendix.</strong></em> I'm trying to achieve this look using square aluminium tubes.</p>
<p><a href="https://i.stack.imgur.com/GCDMi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GCDMi.jpg" alt="enter image description here" /></a></p>
</sub>
| |aluminum| | <p>The only reliable way to make a tube-to-tube connection is using a "thru-bolt", it is removable but aesthetically unpleasant. "Blind bolts" can be considered for the tube-plate connection, but you might damage the base material while trying to remove it.</p>
| 47750 | Reliable non-welding ways to attach square aluminium tubes to aluminium frame? |
2021-10-16T15:36:23.627 | <p>I tore apart a Corsair PC Case Fan, and after stripping away plastic, I found this bearing inside. It has swirly grooves. What are these swirly grooves for? My guess is it has something to do with gripping the black plastic, which seems to be insert molded around the bearing and the motor iron.</p>
<p><a href="https://i.stack.imgur.com/Ve0Ke.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ve0Ke.jpg" alt="Bearing with swirly grooves" /></a></p>
| |motors| | <p>I think this is a variation of the gas dynamic bearing. Such bearings have a long service life due to the fact that they do not have contact between the rotating and static parts. Air passing through special slots raises the rotating part and prevents it from contacting the stationary one.</p>
| 47751 | Why are there grooves on this bearing? |
2021-10-17T02:05:13.860 | <p>I am trying to simulate the following code in NC Viewer (<a href="https://ncviewer.com/" rel="nofollow noreferrer">https://ncviewer.com/</a>).</p>
<pre>O001
N10 G20 G90
N20 M06 T1
N30 M03 S500
N40 M08
N50 G01 X0.25 Y0.25 Z0.25
N60 G00 Z-0.25 F2
N70 G01 X0.4 Y1.75
N80 G01 X0.8 Y0.25
N90 G00 X0.3 Y0.8 Z0.25
N100 G00 Z-.25
N110 G01 X0.65 Y0.8</pre>
<p>But when I simulate (by pressing the icon highlighted in the image below) it the tool moves rapidly. Is there any way to slow it down?</p>
<p><a href="https://i.stack.imgur.com/z4HZ1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z4HZ1.png" alt="enter image description here" /></a></p>
| |simulation| | <p>The double arrows on the far left and far right increase or decrease the speed at which it steps through the code. If you click them it enough times it will actually start to go in reverse.</p>
<p>It seems a bit buggy though and doesn't always have the same behaviour. Might need to press stop then play or just refresh the page.</p>
| 47759 | G code simulation in NC Viewer for Manufacturing Engineering |
2021-10-17T17:51:31.927 | <p><strong>The Problem</strong></p>
<p>I have the following system for which I would like to determine its dynamic response, taken from <a href="https://engweb.swan.ac.uk/%7Eadhikaris/TeachingPages/DampedVibration.pdf" rel="nofollow noreferrer">these lecture notes</a> pg.20:</p>
<p><a href="https://i.stack.imgur.com/Drrqf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Drrqf.png" alt="spring-mass_system" /></a></p>
<p>where the forcing functions are <span class="math-container">$u_1 = 1-U(t - t_0)$</span>, <span class="math-container">$u_2=0$</span>, and <span class="math-container">$u_3=0$</span>, <span class="math-container">$U(\cdot)$</span> is the unit step function and <span class="math-container">$t_0 = 2\pi/\omega_1$</span> where <span class="math-container">$\omega_1$</span> is the first undamped natural frequency.</p>
<hr />
<p><strong>My Attempt</strong></p>
<p>I know the time domain representation of the system to be of the typical format as shown in Eq.(1), and therefore the Fourier transform is as shown in Eq.(2):</p>
<p><span class="math-container">\begin{gather}
\mathbf{M}\ddot{\mathbf{x}} + \mathbf{C}\dot{\mathbf{x}} + \mathbf{Kx} = \mathbf{u}(t) \tag{1} \\
%
\Rightarrow [-\omega^2\mathbf{M} + i\omega\mathbf{C} + \mathbf{K}]\mathbf{X}(i\omega) = \mathbf{U}(i\omega) \tag{2}
\end{gather}</span></p>
<p>where since we know that <span class="math-container">$\mathcal{L}\{1 - U(t - t_0)\} = 1 - \frac{\exp{(-st_0})}{s}$</span>, then the Fourier transform of the right hand side of Eq.(2) is:</p>
<p><span class="math-container">$$\mathbf{U}(i\omega) = \left[\left(1 - \frac{e^{-i\omega t_0}}{i\omega}\right),\, 0,\, 0,\right]^T$$</span></p>
<p>As a result, the dynamic response would be:</p>
<p><span class="math-container">$$\mathbf{X}(i\omega) = [-\omega^2\mathbf{M} + i\omega\mathbf{C} + \mathbf{K}]^{-1}\mathbf{U}(i\omega) \tag{3}$$</span></p>
<p>Since the response by which I am comparing my answer is given in generalised modal coordinates <span class="math-container">$\mathbf{q}$</span>, then the modal transformation of Eq.(3) follows where <span class="math-container">$\mathbf{V}$</span> is the eigenvector matrix thus we know that the modal mass, stiffness, damping matrices should be <span class="math-container">$\bar{\mathbf{M}} = \mathbf{I}$</span> (i.e. the identity matrix), <span class="math-container">$\bar{\mathbf{K}} = \mathbf{\Omega}^2$</span> (i.e. a diagonal marix containing the square of the natural frequencies), and <span class="math-container">$\bar{\mathbf{C}} = \mathbf{V}^T\mathbf{C}\mathbf{V}$</span> respectively. So Eq.(3) becomes:</p>
<p><span class="math-container">$$\mathbf{q}(i\omega) = \left[-\omega^2\mathbf{I} + i\omega\bar{\mathbf{C}} + \mathbf{\Omega}^2 \right]^{-1}\left[\mathbf{V}^T\mathbf{U}(i\omega)\right] \tag{4}$$</span></p>
<p>Thefore, the dynamic response would be the magnitude of the imaginary numbers obtained from Eq.(4) above, for all positive <span class="math-container">$\omega$</span>.</p>
<p>However, when I follow the above steps and solve the problem in <code>MATLAB</code>, the answer is far from the one provided in the excercise. Admitedly, the code is very simple as well (see below) so I struggle to see where I could have gone wrong:</p>
<pre><code>clear ; clc
% Input
n = 3 ; % Number of DOFs
m = 1 ; % Mass [kg]
k = 1 ; % Stiffnes [N/m]
c = 0.2 ; % Viscous damping [Ns/m]
nv = [1 0 0]' ; % Load application vector (load applied to 1st DOF)
% Matrices
M = diag(m*ones(n, 1)) ;
K = diag(2*k*ones(n,1)) - diag(k*ones(n-1,1),-1) - diag(k*ones(n-1,1),1) ;
C = diag(c*ones(n,1)) ;
% Undamped Natural Frequencies & Modes
[V, Omsq] = eig(K, M) ;
Om = diag(sqrt(abs(Omsq))) ; % Natural Frequencies
% Frequency Aanalysis
t0 = 2*pi/Om(1) ; % Time where unit step is applied
omega = linspace(0, 3, 100) ; % Frequency vector
for i = 1:numel(omega)
s = 1i*omega(i) ;
Us = V'*((1 - exp(-s*t0)/s)*nv) ;
FRF(:,i) = (s^2*eye(n) + s*(V'*C*V) + Omsq)\Us ;
end
% Plotting
figure ; plot(omega, abs(FRF)) ; grid on ; set(gca, 'YScale', 'log') ; ylim([0 10]) ;
</code></pre>
<p><a href="https://i.stack.imgur.com/We0dG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/We0dG.png" alt="enter image description here" /></a></p>
| |transfer-function|frequency-response|homework| | <p>As TimWescott indicated, this problem is more suitable for <strong>control theory</strong> for a fuller understanding than for digital signal processing, yet there's no <em>control.se</em> as far as I know. Hence you can give it a try here too since such damped-spring systems is not uncommon subject for audio / acoustic engineers etc.</p>
<p>First of all you have a programming error: Your for-loop variable is set to <strong>i</strong> , a potential error because there's already the <strong>complex unit</strong> <span class="math-container">$j$</span> in your body code in the line <span class="math-container">$$s = j \omega \tag{1}$$</span>which you have implemented as :</p>
<p><code> s = 1i*omega(i) </code></p>
<p>which is wrong: you are using the same letter <strong>i</strong> , both for indexing integer into the angular frequency vector <strong>omega</strong>, and also for the imaginary unit <span class="math-container">$j$</span>. I've replaced the for-loop variable with <strong>k</strong> instead.</p>
<p>Second, you have a signal processing mistake. Your nonzero input function is <span class="math-container">$$ u_1(t) = 1 - u(t-t_0) \tag{2}$$</span>whose (bilateral - double sided) Laplace transform is</p>
<p><span class="math-container">$$ - \frac{ e^{ -s t_0} }{s} \tag{3} $$</span></p>
<p>You can see this by noting that the input is actually: <span class="math-container">$$u_1(t) = u(t_0 - t) \tag{4}$$</span></p>
<p>And the third issue is about the diagonalized computation. I'have replaced the diagonalised frequency response calculation, with the undiagonalised one with the matrcices <strong>M</strong>,<strong>K</strong>, <strong>C</strong>. The result is improved,as shown on the plot below, but still not exactly what you have been given as the answer plot. I guess you may correct it further.</p>
<p>I put the corrected code and the corresponding MATLAB output below :</p>
<pre><code>clc; clear variables ; close all
% Input
n = 3 ; % Number of DOFs
m = 1 ; % Mass [kg]
k = 1 ; % Stiffnes [N/m]
c = 0.2 ; % Viscous damping [Ns/m]
nv = [1 0 0]' ; % Load application vector (load applied to 1st DOF)
% Matrices
M = diag(m*ones(n, 1)) ;
K = diag(2*k*ones(n,1)) - diag(k*ones(n-1,1),-1) - diag(k*ones(n-1,1),1) ;
C = diag(c*ones(n,1)) ;
% Undamped Natural Frequencies & Modes
[V, Omsq] = eig(K,M);
Om = diag(sqrt(abs(Omsq))) ; % Natural Frequencies
% Frequency Aanalysis
t0 = 2*pi/Om(1) ; % Time where unit step is applied
omega = linspace(0, 2.5, 100) ; % Frequency vector
for k = 1:numel(omega)
s = 1i*omega(k) ;
Us = -(exp(-s*t0)/s)*nv;
FRF(:,k) = (s^2*M + s*C + K)\Us ;
%Us = V'*((1 - exp(-s*t0)/s)*nv) ;
%FRF(:,k) = (s^2*eye(n) + s*(V'*C*V) + Omsq)\Us ;
end
% Plotting
figure ; plot(omega, abs(FRF)) ; grid on ; set(gca, 'YScale', 'log') ; ylim([0.01 10]) ;
</code></pre>
<p><a href="https://i.stack.imgur.com/ZiXUB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZiXUB.png" alt="enter image description here" /></a></p>
| 47765 | How to determine the dynamic response of a spring-mass system? |
2021-10-18T18:17:22.040 | <p>How would I make a drawing similar to this ? Would I have to use Creo parametric draw a sketch and then convert this into a drawing ?</p>
<p>Or would I need to use a different Creo software ?</p>
<p><a href="https://i.stack.imgur.com/zRI5j.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zRI5j.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|manufacturing-engineering| | <p>What you suggest is fine, I've done this at occasion but I would prefer not to*. However, bear in mind that this is a manufacturing detail, not a design detail.</p>
<p>In general Creo is meant for making design details and manufacturing details should be handled by a CAM module downstream. Why? Well the design usually has a longer time span than the manufacturing detail, and manufacturing floor might want to interleave it with different jobs at different times and optimize for different needs.</p>
<p>* There are other strategies that you can employ instead. Just design the object you are building. Then make:</p>
<ol>
<li>A secondary assembly that lays out the components and make a drawing out of that.</li>
<li>If all the parts go to the same design you may do this layout as a explode state.</li>
<li>Let the CAM module lay them out (this is best as it gives manufacturing much more flexibility, also most likely it can be auto optimized)</li>
</ol>
<p>the benefit of this is that the instructions are associative with the actual designs and not a self contained drawing</p>
| 47780 | Creating a drawing in Creo |
2021-10-18T18:38:02.600 | <p>I'm looking for a hinge</p>
<ul>
<li>that can open a maximum of 90 degrees</li>
<li>will open fully by gravity alone</li>
<li>approx 20mm wide</li>
<li>approx 50mm long each arm</li>
<li>approx 10mm thick when closed</li>
</ul>
<p>Is this something that can be sourced?
Could it be 3D printed?</p>
| |hinge| | <p>Here is a full page of 90 degrees "stop" hinges for cabinets and doors. You have to browse through to find one that fits your need.</p>
<p><a href="https://i.stack.imgur.com/lt9jx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lt9jx.png" alt="enter image description here" /></a></p>
<p><a href="https://www.amazon.com/90-degree-stop-hinge/s?k=90+degree+stop+hinge" rel="nofollow noreferrer">https://www.amazon.com/90-degree-stop-hinge/s?k=90+degree+stop+hinge</a></p>
| 47781 | Hinge type 90degree max |
2021-10-19T02:44:48.547 | <p>I am developing Five-in-One machine tool which I did as a project working during 1984-85 in our Polytechnic college. Even though we made prototype at that time, now I want to make prototype for production version. It can be a cubical size of 1-1/2 to 2 meters.</p>
| |machine-design|machine-elements| | <p>IMHO the most important parameter that will affect the thickness are the vibrational performance of the machine bed, which in turn depends on :</p>
<ul>
<li>rpms of the motors</li>
<li>feed rates</li>
<li>load</li>
<li>support of the bed</li>
<li>etc.</li>
</ul>
<p>Probably you need to make it thick enough to be:
a) rigid
b) with a large mass.</p>
| 47786 | What should be the thickness of machine tool bed? |
2021-10-20T00:46:38.043 | <p>This topic is totally alien to me, so I may be asking something very obvious.</p>
<p>I am designing a cooling system. In the bulk of aluminum, I drill 3mm holes and then I connect them one to another to create a long path for the stream.</p>
<p>The question is now, how long can this maze be? What I am trying to figure out to answer this question is first of all, what pressure is required to drive water into this pipeline. And to determine that, I believe, somewhere should be a coefficient of friction between water and aluminum. Though I can't seem to find a useful chart, it's all about solid bodies friction. Is it because the coefficient might in fact depend on the pipe diameter?</p>
| |pressure|friction|cooling|pipelines|liquid| | <p>The terms you are looking for are Major and Minor pressure losses. Which are a misnomer for systems on the scale your are talking about, since "Minor" losses are usually the greater. The pressure drop across any feature of the system (sudden expansion, sudden contraction, a split, a component like a valve) is:</p>
<p><span class="math-container">$\Delta P = 0.5 k\rho Q^2/A^2$</span></p>
<p><span class="math-container">$\rho$</span> density, <span class="math-container">$Q$</span> volumetric flow rate, <span class="math-container">$A$</span> cross sectional area. <span class="math-container">$k$</span> is a factor you can look up in a table for Minor losses. Valves and filters also have published <span class="math-container">$C_V$</span> values, which are used in a different formula to relate pressure drop to flow rate. Major losses are the friction in a straight pipe, and <span class="math-container">$k=(f*L/D)$</span> is expressed as a function of friction factor (very conservatively <span class="math-container">$f=0.1$</span>), length, and diameter. For non-civil engineering flow systems, major losses are often negligible.</p>
<p>Add all these Delta Ps together across the system for the required flow rate, and you can size your pump. Often with a healthy factor, like double.</p>
| 47796 | Water dynamics in aluminum pipeline for dummies |
2021-10-20T13:23:15.767 | <p>I have the following question:</p>
<p><a href="https://i.stack.imgur.com/sdaPQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sdaPQ.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/CoWoz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CoWoz.jpg" alt="enter image description here" /></a></p>
<p>Here is my attempt:</p>
<p><a href="https://i.stack.imgur.com/xXz5b.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xXz5b.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/cMhDH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cMhDH.jpg" alt="enter image description here" /></a></p>
<p>Is this correct ?</p>
| |mechanical-engineering|structural-engineering|structural-analysis|stresses| | <p>You made a mistake in solving the reaction at joint "A". See calc below:</p>
<p><a href="https://i.stack.imgur.com/emLqU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/emLqU.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$\sum M_G = 0$</span></p>
<p><span class="math-container">$R_A = \dfrac{22.31*8}{12} = 14.873$</span> kN</p>
<p>Solve internal member force using the method of section:</p>
<p><a href="https://i.stack.imgur.com/PIPd5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PIPd5.png" alt="enter image description here" /></a></p>
<p>Since there is only one unknown in the vertical direction, so we can solve the member force <span class="math-container">$F_{BC}$</span> directly by <span class="math-container">$\sum F_X = 0$</span></p>
<p><span class="math-container">$\sum F_X = 0$</span></p>
<p><span class="math-container">$-F_{BC}cos 30^{o}$</span> + R_A = 0</p>
<p><span class="math-container">$F_{BC} = \dfrac{R_A}{cos 30^{o}} = \dfrac{14.873}{0.866} = 17.17$</span> kN (Tension. Direction as assumed - away from the joint B)</p>
<p><span class="math-container">$\sigma_{BC}$</span> = <span class="math-container">$\dfrac{17.17}{0.08} = 214.6 kN/m^2 = 214.6 kPa = 214,600 Pa$</span></p>
| 47798 | Calculating member stress in a truss |
2021-10-21T16:56:59.263 | <p>The maximum shear stress which we get from the equation: <span class="math-container">$\frac{VQ}{It}$</span> (for instance, the maximum shear stress at the neutral axis for a rectangular beam cross section, fixed at one end and tranverse loaded on the other can be thought of as: <span class="math-container">$\frac{3V}{2A}$</span>), while the maximum shear stress which we get from the Mohr's Circle (i.e. <span class="math-container">$\frac{principal_{max} - principal_{min}}{2}$</span>), so what is the difference between these two values? These both are maximum values of the shear as learned at the university level, so it is slightly confusing. Is former absolute while the latter is maximum?</p>
<p>Are these two shear stresses supposed to occur at the same location or different?</p>
| |mechanical-engineering|solid-mechanics|shear| | <p><span class="math-container">$f = \dfrac {VQ}{Ib}$</span> is the "shear flow" in the flexural beam element caused by the applied load, and its intensity varies along the beam as the internal shear force <span class="math-container">$V$</span> varies.</p>
<p><a href="https://i.stack.imgur.com/x8Eoh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/x8Eoh.png" alt="enter image description here" /></a></p>
<ul>
<li>The shear stress is always starting from zero at the free surface because shear occurs at the interface of sliding elements as shown below:</li>
</ul>
<p><a href="https://i.stack.imgur.com/OLsew.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OLsew.png" alt="enter image description here" /></a></p>
<p>Mohr's Circle is used to find the stresses on an inclined plane of an axially loaded member as depicted below:</p>
<p><a href="https://i.stack.imgur.com/zozh8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zozh8.png" alt="enter image description here" /></a></p>
<p><strong>So, "No", the two types of shear stress (due to flexural and due to axial load) are completely different matters.</strong></p>
| 47810 | What is the difference between max shear stress at neutral axis in bending and max shear stress from Mohr's Circle? |
2021-10-21T23:39:18.700 | <p>I have a 4"x10" wood beam that I want to hang a swing from, but the top of it is not accessible (the roof is above).</p>
<p>Would it be more stable/load bearing to do a shoulder eye bolt with a nut horizontally through the beam, with the load being at 90 degrees, or using an eye bolt from below that's screwed into the wood?</p>
<p>I figured the horizontal lag bolt can have basically arbitrary working load depending on the diameter, but i wasn't sure about what the 90 degree load would do to the beam.</p>
<p>Thanks!</p>
| |structural-engineering| | <p>I would use a 1/8 x 4 x 10-inch plate on each side with a proper pre-drilled hole on it for your eye bolt with four 1/4 inch bolts on each corner.</p>
<p>A swing not supported by a plate will eventually split the beam.
Hanging them to the bottom of the beam has its own risks of failure.</p>
| 47822 | How to hang weight from beam |
2021-10-23T13:04:00.947 | <p>Consider a block of dimensions <span class="math-container">$l, b, h$</span> kept on a frictionless surface, initially at some uniform temperature <span class="math-container">$ T_i $</span>. Assume the material to be homogeneous and isotropic.</p>
<p>When the block is heated uniformly there are no stresses developed in the block because the block is not restrained to move. It can expand in all directions freely.</p>
<p>When the block is heated non uniformly, my textbook says there will be stresses developed. Why is that so? The block is still not restrained to move.</p>
| |mechanical-engineering|structural-analysis|thermal-expansion| | <p>Imagin the block is heated on the bottom and has expanded to <span class="math-container">$l+\delta l$</span>, but this expansion tries to expand the top part along with it.</p>
<p>So the will be some strain between the hot lower part and the cold top part which will cause stress until the whole block is at temperature <span class="math-container">$T_i+\delta i$</span></p>
<p>Some years ago I had to supervise a steel frame structure. They had stacked steel columns on the site where they would be exposed to the direct sun at around 11 am daily.</p>
<p>When the sun hit the stack of columns the top ones would expand and move the lower rows with them creating a loud screeching sound. As the project progressed and lower-level columns were lifted by the crane, some were damaged and had crushed spots on the flanges due to the thermal action of the top layer.</p>
| 47850 | Why stresses develop during non-uniform heating and not during uniform? |
2021-10-24T11:51:16.037 | <p>Consider a bar which is fixed at one of the ends. I increase the temperature of the bar by <span class="math-container">$\Delta T$</span>. The sources I'm referring to say that stress developed in the bar will be zero because the bar is free to expand.</p>
<p>However the bar is free to expand only in the longitudinal direction not in the lateral ones. So how the stress can be zero.</p>
<p>If the bar would've rested on a frictionless surface then the stress would've been zero. But how in this case?<a href="https://i.stack.imgur.com/rt8sQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rt8sQ.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-analysis|stresses|thermal-expansion| | <p>In practical engineering, a "bar" is something which only transmits axial and (maybe) torsional loads.</p>
<p>You are right that there are some local stresses where the bar is fixed to the plate. That is true however the bar was fixed (e.g. welded, bolted, etc). The local stresses are self-equilibrating.</p>
<p>If you <em>care</em> about the local stresses, you don't model the object as a bar. You model it in full 3-dimensional detail - probably including the nonlinear effects of geometric tolerances in the connection, the pre-stress caused by torque and local plastic deformation in bolted joints, etc.</p>
| 47857 | Bar fixed at one end at heated |
2021-10-25T00:23:02.447 | <p>Could we reuse expired wells in places like Huntington beach California, to desalinate water? I found the following chart online which suggests that some deeper wells could have temperatures high enough to evaporate water.</p>
<p><a href="https://i.stack.imgur.com/IsxFi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IsxFi.png" alt="enter image description here" /></a></p>
<p>I am seeing that at 4000 meters, the Earths crust is comfortably 100 degrees Celsius in the 'worst' area drilled for this data. Is there any obvious reason that I wouldn't be able to retrofit an old well into a geothermal desalination plant? Safety is my primary concern, both in operating the retrofitted well without collapse, as well as extracting healthy water from the well. Second concern would be regarding how much water can be boiled at a time without the temperature of the well falling below 100C. Third concern would be what machinery and structures I would need to install in order to get steam out while water goes in.</p>
<p>Please poke holes into this idea.</p>
| |thermodynamics|pumps| | <p>It's not quite as simple as you are imagining it to be, but using geothermal heat as part of a multi-effect distillation plant or other type of plant is proven tech. But the prime use is for preheat. All the fun stuff still happens on the surface with vacuum distillation and membrane systems working on the preheated brackish water. For seawater desalination, you basically need to park the desalination plant next to a powerplant and use higher quality waste heat (a lot of it). But for mildly brackish water, geothermal preheat can be adequate and plonking down a plant near accessible geothermal sources is perfectly reasonable.</p>
<p>The simple system would to extract hot mineral water from a geothermal water well and immediately purify it. There are more dissolved minerals in hot water, but it can still work out depending on what they are. But you can also run a heat exchanger loop to transfer the heat to a better water supply if available. The geothermal heat can be put to use in myriad ways, with temperature-driven membrane purification probably the most promising. But you also need a cold side for this to work well.</p>
<p>Of course, if you are sitting on top of a wicked good geothermal source you can handle about anything.</p>
<blockquote>
<p>In the first phase of the Aqua Genesis plan to remedy the Salton Sea, a complex of Firestone's units would be assembled at the Sea's Bombay Beach near Niland, California. (Eventually the company plans to produce over 5000 of the units at a factory in Nevada.) The plant would siphon Salton Sea water, which has a percentage point higher salinity level than ocean water's 3.5 percent. The water would then be heated with geothermal energy drawn from 4700 feet below ground. Geologists, according to Firestone, know water in the area at that depth to be 656 degrees F. Once its heat has been taken and its pressure used to move Salton Sea water through the plant, the operation would return the groundwater to its source.</p>
<p><a href="https://www.sandiegoreader.com/news/2004/feb/26/inventor-tackles-salton-sea-disaster/" rel="nofollow noreferrer">https://www.sandiegoreader.com/news/2004/feb/26/inventor-tackles-salton-sea-disaster/</a></p>
</blockquote>
<p>PS, we expect a modicum of research before asking here, and that might include discovering the Wikipedia article on <a href="https://en.wikipedia.org/wiki/Geothermal_desalination" rel="nofollow noreferrer">Geothermal Desalination</a>. Its not a great read, but it would lead you to other sources of information.</p>
| 47867 | Reusing Old Wells to Desalinate Water with Geothermal |
2021-10-25T09:19:41.583 | <p>As far as I understood <a href="https://en.wikipedia.org/wiki/Heating_element" rel="nofollow noreferrer">heating elements</a> make use out of <a href="https://en.wikipedia.org/wiki/Joule_heating" rel="nofollow noreferrer">Joule heating</a>. Hence, I wonder why one would preferably use materials with a negative temperature coefficient, therefore?
Isn't it, that, the higher the resistance, the more warmth will be generated? So why would you want that the resistance decreases with temperature?</p>
| |heat-transfer|heating-systems|electrical|temperature| | <p>It really depends. If you have multiple heating elements (or areas within a single element), in parallel, they will end up at slightly different temperatures. If the application is temperature limited, then the max total system power is determined by the "weakest-link" -- i.e. the hottest individual element.</p>
<p>If the elements are PTC, then hotter elements' increasing R automatically reduces their share of the overall power. It tends to equalize the per-element power. This is generally good.</p>
<p>If the elements are NTC, then the hotter elements' decreasing R will make them draw relatively more current, making them hotter, and this can make the temperature variation between elements more extreme, thus in effect lowering the overall system power that is allowed.</p>
| 47871 | Why materials with preferably a negative temperature coefficient for heating elements? |
2021-10-25T20:55:03.277 | <p>I was going through the basic assumptions made by Bernoulli beam theory. I realized that one of the most important assumption in this theory is that the deformations and rotations of the beams are supposed to be small, not large. I couldn't understand that how is this important and if it could invalidate the Bernoulli beam theory if they are not small? In some places, its written that it is to make the calculations easier and reach to a more understandable form, while some sites write that it is because if the deformations/rotations are large then significant shear deformations develop on the cross section. I don't know what is the actual reason behind it. I mean even if the shear deformations develop, then isn't it possible to super impose the deformations from Bernoulli beam equation onto the ones coming from shear? Or does it completely invalidates the Bernoulli beam theory results if deformations are large?</p>
<p>Plus, how would we decide that this deformation is small and this deformation is large?</p>
| |beam|solid-mechanics| | <p>Apart from the other answers another reason why the Bernoulli-Euler beam theory falls apart at large deformation is due to the approximation about the radius of curvature.</p>
<p>The Bernoulli-Euler beam theory is usually encountered in the following equation form:</p>
<p><span class="math-container">$$EI w''(x) = M(x)\qquad\text{ or } \qquad w''(x) = \frac{M(x)}{EI }$$</span>
where:</p>
<ul>
<li>E: elastic modulus</li>
<li>I: second moment of area of the crosssection</li>
<li>w(x): the function of deflection of the beam along its length (<span class="math-container">$w''(x)$</span> is its second derivative wrt x)</li>
<li>M(x) : the function of the bending moment along the length of the beam x</li>
</ul>
<p>However, <span class="math-container">$w''(x)$</span> is an approximation of the curvature of the beam <span class="math-container">$\kappa$</span> (and it's inverse, the radius of curvature <span class="math-container">$\rho$</span>). i.e. the equation is normally:</p>
<p><span class="math-container">$$ \kappa(x) = \frac{1}{\rho}= \frac{M(x)}{EI}$$</span></p>
<p>However, the curvature of a function f(x) is given by:</p>
<p><span class="math-container">$$\kappa(x) = \frac{|f''(x)|}{(1+f'(x)^2)^{\frac{3}{2}}}$$</span></p>
<p>This is where the assumption of is important. For small deformations the <span class="math-container">$f'(x)$</span> is small, so <span class="math-container">$f'(x)^2)$</span> is even smaller, therefore it can be neglected. So:</p>
<p><span class="math-container">$$\kappa(x) = \frac{|f''(x)|}{(1+f'(x)^2)^{\frac{3}{2}}}\rightarrow \kappa(x)=\frac{|f''(x)|}{(1+0)^{\frac{3}{2}}}\rightarrow \kappa(x)=\frac{|f''(x)|}{1}$$</span></p>
<p>and therefore:</p>
<p><span class="math-container">$$\kappa(x)=|f''(x)|$$</span></p>
<p>for large displacements the above simplification cannot be assumed.</p>
| 47882 | Why shouldn't we use Bernoulli beam theory if deformations and rotations are large? |
2021-10-26T16:21:27.673 | <p>Is it possible to open a part created using Creo commercial edition in the student edition ? I tried this but it doesn’t seem to work. Is there a way to do this ?</p>
| |mechanical-engineering|design|cad| | <p>The "professional" version has more features, so the parts have extra information that the student version won't be able to read or convert.</p>
<p>So, You will need the "professional" version - at least for complicated parts. If it was just a simple bar it would likely open.</p>
<p>You can understand them protecting their software...</p>
| 47903 | Opening Creo file in student edition |
2021-10-26T18:18:25.847 | <p>I recently received an order of extruded aluminum strip 3/32" thick that supposed to be 6061 hardened to specification ASTM B221. However, when we tried to bend it, it broke and the interior looked like cast zinc coated with aluminum. Is 6061 supposed to look that way and be unable to bend 90-degrees?</p>
<p>How can I verify that I have solid aluminum and not some fake alloy with zinc in it?</p>
| |metallurgy| | <p>The standard simple test is to do vickers hardness testing, at a fixed load. the HV should be specified for the material and condition. I'm assuming you are using something like 6061 in the T6 condition. Random online papers suggest that the hardness should be 95HV (load not specified).</p>
<p>Your density test won't tell you too much about the Al grade, as the alloying elements are usually dilute, so don't change the overall weight to more than a few %.</p>
<p>The more complicated ways to confirm the material is to do either XRF, EDS or wet-chemical testing (in order of increasing cost), which can be used to confirm the chemistry, but not the mechanical properties. XRF and EDS are quick to do, but all these things require specialised equipment.</p>
<p>When you say that it wont bend 90*, you've not specified the important parameter, which is the bend-radius, and how it was bent. A two point bend-test and a three point bend-test will give different results for cracking, as will whether you do it at low or high temperature. If you do it at high temperature for any period of time (>150C) then you will likely invalidate the material condition by forcing ageing, and internal precipitation.</p>
| 47905 | How can I tell if I got counterfeit aluminum? |
2021-10-27T09:41:05.280 | <p>This is my first question / post here.
I have just begun researching motors and how they work, as they are necessary for a personal project.<br/>
I want to know how to control a motor so that it only rotates 45 degrees forwards and backwards. I would be implementing this by rotating a pole bi-directionally, with a fixed starting position. Any help would be appreciated.</p>
| |mechanical-engineering|electrical-engineering|motors| | <p>(Remote Control) RC servos usually have that type of functionality and they --usually-- use a PWM pulse to control the position.</p>
<p>The standard idea is (used to be) that you provide a digital pulse (either Hi or Low) 50 times per second (so the duration of the pulse is 20[ms]). Modern RC servos can be more forgiving and can have different times.</p>
<p>The duration of the high pulse is essentially the signal that controls the position. Typically for a 20ms duration, the central position is at 1.5ms (or 7.5 duty cycle).</p>
<p><a href="https://i.stack.imgur.com/NcoNG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NcoNG.png" alt="enter image description here" /></a></p>
<p><strong>figure: <a href="https://en.wikipedia.org/wiki/Servo_control#/media/File:Servomotor_Timing_Diagram.svg" rel="nofollow noreferrer">wikipedia</a></strong></p>
| 47916 | Controlling Motor Rotation |
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