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2021-12-23T06:55:47.063 | <p>What are disadvantages if I increase the gain Kd for a PD(Proportional Plus Derivative) controller, someone told me larger gain for derivative error for will reduce the stability margin, while the text books said increase the Kd could increase the damping ratio and high damping ratio means more stability margin.</p>
| |control-engineering|control-theory| | <p>Depending on the plant, a reduction in phase margin could happen. Here is a theoretical example. Consider a plant with a double integrator and an additional pole:</p>
<p><span class="math-container">$$\frac{K_1}{s^2(1+s/\omega_P)}$$</span></p>
<p>Now let's have a PD controller, which could be written either
<span class="math-container">$[K_P(1+{K_D}s)]$</span> , or as <span class="math-container">$[K_P + {K_D}s]$</span> . For convenience, both of these could be re-written in the form</p>
<p><span class="math-container">$$K_0(1+s/\omega_Z)$$</span></p>
<p>The loop's stability can then be analyzed by looking at the product of the gains going around the loop -- i.e. controller and plant. We have two poles at the origin, a zero at <span class="math-container">$\omega_Z$</span> and a pole at <span class="math-container">$\omega_P$</span>.</p>
<p>An hand-drawn Bode plot is below, followed by an Octave plot of essentially the same thing. Each graph shows two different values of <span class="math-container">$\omega_Z$</span>, corresponding to two different values of <span class="math-container">$K_D$</span>.</p>
<p><a href="https://i.stack.imgur.com/aDVyn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aDVyn.jpg" alt="hand drawn" /></a>
<a href="https://i.stack.imgur.com/rPSHV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rPSHV.png" alt="auto-loopgain-bode" /></a></p>
<p>For completeness, here are the corresponding closed loop bode plots and closed loop step responses</p>
<p><a href="https://i.stack.imgur.com/5bvqK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5bvqK.png" alt="auto-cl-bode" /></a>
<a href="https://i.stack.imgur.com/XLNI8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XLNI8.png" alt="auto-cl-step" /></a></p>
<p>Hopefully it's possible to see from first bode plots (the ones looking at around-the-loop), that reducing <span class="math-container">$\omega_Z$</span> (i.e. increased <span class="math-container">$K_D$</span>), by 10x in this example, moves the first breakpoint left by 10x. This then lifts the section of the line with the -20dB/decade slope up by 20dB, causing it to intersect the horizontal axis about 10x further to the right, meaning the loop has about a 10x higher bandwidth.</p>
<p>The movement of the gain crossover frequency also results in a different phase margin. In the example shown, it reduces phase margin from something like 80 degrees for line (1), to 50 degrees for line (2).</p>
<p>However, if the overall gain were smaller, the opposite could also be the case -- if the gain crossover started out to the left of the "peak" in the phase plot, then the phase margin would increase at first, but then decrease again after the gaincrossover frequency passes the peak in the phase plot.</p>
| 48871 | What are disadvantages if increasing the gain Kd for a PD(Proportional Plus Derivative ) controller |
2021-12-24T16:43:29.830 | <p><a href="https://i.stack.imgur.com/Bt5EB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Bt5EB.jpg" alt="enter image description here" /></a></p>
<p>These are how circular turning tracks look like. Now , according to my textbook:<a href="https://i.stack.imgur.com/qI9qO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qI9qO.jpg" alt="enter image description here" /></a></p>
<p>This is how it looks like. But I argue upon each & every point of it.</p>
<ol>
<li>Here , the car is turning but it’s direction of WHEELS is straight , which implies that direction of velocity of car is tangentially straight. How is that possible ? If the car is turning , then the front wheels of car also turn & are not straight in direction. Only the direction of back side wheels of car is straight.</li>
</ol>
<p>Even if I think for a time dt covered by car , then also car is going to turn in direction which I think is VT but not straight. Where am I going wrong ?</p>
<p>EDIT: I was revising Acceleration in circular motion & I got to know that a particle moving in a circle has 2 components. Radial & tangential vectors. We know if there are two component vector , then they have to be of s Resultant vector i.e vector in between the two vectors. So , I think the V in the diagram above has to be tangential velocity & centripetal velocity is just not shown in the diagram. Am I right ?
<a href="https://i.stack.imgur.com/N3oKW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N3oKW.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|automotive-engineering|applied-mechanics| | <p>The car steering is designed to set the alignment of the four wheels to follow the curve of rotation without any skidding.</p>
<p>At any <span class="math-container">$\Delta t$</span> time the car experiences two components of force, tangential<span class="math-container">$F_t$</span> and centripetal <span class="math-container">$F_{centripetal}=mv^2/r$</span></p>
<p>If the car is not accelerating tangentially, the <span class="math-container">$F_t=0$</span> and the only force acting on the car is <span class="math-container">$F_s,$</span> which is equal and opposite of the centripetal force as your diagram.</p>
<p>'</p>
<p><a href="https://i.stack.imgur.com/rVZUu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rVZUu.png" alt="eturn" /></a></p>
| 48890 | Why is direction of velocity in front direction |
2021-12-24T18:41:43.383 | <p>Recently, I dipped my spoon into my tea. I saw that as I dipped my spoon into my tea, two vortices formed at both edges of the spoon. I guess this is similar to vortices forming at the end of aeroplane wings. Here I assume both vortices have equal strength and according to Kelvin's circulation theorem, the sum of circulation must equal zero in my system (my cup of tea).</p>
<p>However, I noticed that if I stirred my tea first and then dipped my spoon into my tea (whilst the fluid was rotating), two vortices would still form (for a very short period of time) and then they would combine together (since the fluid was still rotating from when I stirred my tea). They would form one vortex. However, I would like to ask why the two vortices don't cancel out? Are they not of equal strength?</p>
| |fluid-mechanics|aerospace-engineering|aerodynamics| | <p>Trying to "see" your experience without pictures.</p>
<p>First, when you inserted your spoon, you could have lowered it at an angle causing it to displace the liquid like a row causing two trailing vortices at the edges.</p>
<p>Later when you dipped it into circulating tea, the current interacted with the two vortices. The current would accentuate and pull in a vortex rotating in the same direction concentrically, and repel the other vortex, hence merging them.</p>
| 48893 | Kelvin's Circulation theorem applied to a cup of tea |
2021-12-25T01:49:34.820 | <p>In some industries, as in biodiesel production, upon reaction, there will be several products with different properties (density) generated inside one chamber, and it's crucial to separate each of them.</p>
<p><a href="https://i.stack.imgur.com/aKSTw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aKSTw.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/jvLWm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jvLWm.png" alt="enter image description here" /></a></p>
<p>In the lab, we can accomplish this using a <a href="https://en.wikipedia.org/wiki/Separatory_funnel" rel="nofollow noreferrer">separation funnel</a></p>
<p>Or <a href="https://en.wikipedia.org/wiki/Decantation" rel="nofollow noreferrer">decantation</a> can also be an alternative when lab-instrument is absent.</p>
<p>But now I wonder how it's done on an industrial scale?</p>
| |industrial-engineering| | <p>I would use</p>
<ul>
<li><p>A tall narrow storage tank to separate layers into long cylinders one on top of the other with several valves at appropriate heights.</p>
</li>
<li><p>Wait for them to settle</p>
</li>
<li><p>Insert a densitometer or install manometers to make sure they are stratified.</p>
</li>
<li><p>Open one valve at a time at the middle height of desired liquid with a controlled flow as to have no turbulence or remixing of different liquids.</p>
</li>
<li><p>Divert the discharge to a second tank for further refinement if needed.</p>
</li>
</ul>
<h2>Edit</h2>
<p>In response to OP's comment as to control of the valves:</p>
<p>There are electrical signal-controlled valves routinely used in industry that can be controlled by a digital densitometer such as the photo.</p>
<p>'</p>
<p><a href="https://i.stack.imgur.com/L5Lnz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/L5Lnz.jpg" alt="elect valve" /></a></p>
| 48899 | How can the function of a separation funnel be accomplished industrially? |
2021-12-25T09:17:17.410 | <p>Problem:</p>
<p><a href="https://i.stack.imgur.com/8FkYs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8FkYs.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/6iy5um.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6iy5um.png" alt="enter image description here" /></a></p>
<hr />
<p>I can understand the problem physically that the block is pulled by some amount X so that its displacement at t=0 is X. It is then released from that position so that its velocity at t=0 is zero. So by the physical nature of the problem I can deduce that the displacement vs time graph will look like:</p>
<p><a href="https://i.stack.imgur.com/eXZVWm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eXZVWm.jpg" alt="enter image description here" /></a></p>
<p>However I'm having trouble understanding the problem from mathematical standpoint. I've learnt that the equation of motion for an underdamped system is given as, <span class="math-container">$x(t)$</span></p>
<p><a href="https://i.stack.imgur.com/47hckm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/47hckm.png" alt="enter image description here" /></a></p>
<p>where <span class="math-container">$X_o$</span> and <span class="math-container">$\phi_o$</span> can be determined from initial conditions. So my understanding tells me that if I apply the boundary conditions as given in the problem I must obtain <span class="math-container">$X_o = X$</span> and <span class="math-container">$\phi_o = \frac{\pi}{2}$</span>.</p>
<p>So I proceeded as follows:</p>
<p><a href="https://i.stack.imgur.com/ERQmpm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ERQmpm.jpg" alt="enter image description here" /></a></p>
<p>Using Boundary Conditions I'm getting</p>
<p><span class="math-container">$$tan\phi_o=\frac{\sqrt{1-\zeta^2}}{\zeta}$$</span></p>
<p>which wont reduce further to <span class="math-container">$\phi=\frac{\pi}{2}$</span></p>
<p>where am I going wrong?</p>
<hr />
<p>P.S. - I'm just having trouble with mathematically coming to the conclusion that <span class="math-container">$X_o = X$</span> and <span class="math-container">$\phi_o = \frac{\pi}{2}$</span>. I will be able to work out for what the problem is actually asking - amplitude after n cycles, by myself.</p>
| |mechanical-engineering|vibration|homework| | <p>I think your math is taking you to the right place. Looking at the this sketch from the problem statement, it does not look like <span class="math-container">$\dot{x}(0) = 0$</span> when <span class="math-container">$\phi_0=\pi/2$</span>; for example <a href="https://www.wolframalpha.com/input/?i=slope+of+%28e%5E%28-t%29%29%28sin%283t%2Bpi%2F2%29%29+at+t%3D0" rel="nofollow noreferrer">see here</a>. Perhaps that may be the source of confusion?</p>
<p><a href="https://i.stack.imgur.com/eXZVWm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eXZVWm.jpg" alt="sketch" /></a></p>
<p>So here's my crack at it. I changed the notation slightly. <span class="math-container">$\omega_0$</span> and <span class="math-container">$\omega_1$</span> would be expressed in terms of <span class="math-container">$\omega_n$</span> and <span class="math-container">$\zeta$</span> . Also used <span class="math-container">$A$</span> for the scale factor, since I don't think <span class="math-container">$X_0$</span> is the same as <span class="math-container">$x(0)$</span>.</p>
<p><a href="https://i.stack.imgur.com/UBulV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UBulV.jpg" alt="maff" /></a></p>
<p>Quick check with wolfram alpha, using A=1, <span class="math-container">$\omega_0$</span>=1, <span class="math-container">$\omega_1$</span>=3: <a href="https://www.wolframalpha.com/input/?i=value+of+%28e%5E%28-t%29%2Fsin%28atan%283%29%29%29%28sin%283t%2Batan%283%29%29%29+at+t%3D0" rel="nofollow noreferrer">value</a> and <a href="https://www.wolframalpha.com/input/?i=slope+of+%28e%5E%28-t%29%2Fsin%28atan%283%29%29%29%28sin%283t%2Batan%283%29%29%29+at+t%3D0" rel="nofollow noreferrer">slope</a>:</p>
| 48905 | A problem from Damped free Vibrations |
2021-12-25T16:46:46.437 | <p>The atmospheric pressure at the surface of the earth is about 101 <span class="math-container">$kPa$</span>.</p>
<p>For Carbon fiber reinforced plastic (70/30 fibre/matrix, unidirectional, along grain):</p>
<ul>
<li>Young's Modulus is: <span class="math-container">$181\ GPa$</span></li>
<li>Density - <span class="math-container">$1.6\ g/m^3$</span></li>
</ul>
<p>With these parameters in mind:</p>
<p>How do we calculate the minimum thickness/weight of a vacuum sealed (i.e. inside the container is vacuum) spherical container/vessel that would withstand the standard atmospheric pressure without much change in shape with its inner volume of <span class="math-container">$k\ m^3$</span>?</p>
<p>And is there a general formula to calculate this, just based on the external surface area (independent of shape of the container)?</p>
<p>My background isn't in Physical Engineering sciences thus an apology beforehand.</p>
| |mechanical-engineering|pressure-vessel| | <p>Typically the maximum stress for a thin walled spherical pressure vessel for a given diameter, thickness and pressure is:</p>
<p><span class="math-container">$$\sigma = \frac{P\cdot R}{2\cdot t}$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$\sigma$</span> is the observed pressure (in Pa)</li>
<li><span class="math-container">$P$</span> is the pressure (in Pa)</li>
<li><span class="math-container">$R$</span> is the radius of the vessel (in m)</li>
<li><span class="math-container">$t$</span> : is the thickness of the vessel (in m)</li>
</ul>
<p>using this it is possible to find the minimum mass of a pressure vessel is (<a href="https://en.wikipedia.org/wiki/Pressure_vessel#Spherical_vessel" rel="nofollow noreferrer">Wikipedia</a>)</p>
<p><span class="math-container">$$M={3 \over 2}PV{\rho \over \sigma }$$</span>,
where:</p>
<ul>
<li>M is mass, (kg)</li>
<li>P is the pressure difference from ambient (the gauge pressure), (Pa) (100000Pa = 1 Atm)</li>
<li>V is volume,</li>
<li><span class="math-container">$\rho$</span> is the density of the pressure vessel material, (expressed in <span class="math-container">$kg/m^3$</span>)</li>
<li><span class="math-container">$\sigma_{all}$</span> is the maximum allowable working stress that material can tolerate. (<span class="math-container">$Pa$</span>)</li>
</ul>
<p>However, an interesting parameter in the question which is without changing much its shape. That can be interpreted as having a very small deformation. Usually a common value which is used to denote small deformation within the elastic range is 0.2% or (0.002). So the value that you could use for the allowable stress is <span class="math-container">$\sigma_{all}= E\cdot 0.002 = 181 GPa \cdot 0.002= 362 MPa$</span></p>
<h2>Compression buckling</h2>
<p>An very important additional check is that of buckling (because the shell will be under compressive loads).</p>
<p>The buckling pressure <span class="math-container">$q_C$</span> of a elastic thin spherical shell was obtained by Zoelly (1915) and Van der Neut (1932):</p>
<p><span class="math-container">$$q_c = \frac{2 E}{[3(1-\nu^2)]^{1/2}} \left(\frac{h}{R}\right)^2 \tag{eq:2}$$</span></p>
<p>where</p>
<ul>
<li>E is the elastic modulus</li>
<li><span class="math-container">$\nu$</span> is the Poisson Ratio</li>
<li><span class="math-container">$h$</span> is the shell thickness</li>
<li><span class="math-container">$R$</span> is the shell radius. Assuming a volume V, the radius of the sphere will be
<span class="math-container">$R= \sqrt[3]{\frac{3V}{4\pi}}$</span></li>
</ul>
<p>Therefore, given that the critical pressure needs to be 1 atm (<span class="math-container">$q_C= 1Atm = 100000Pa$</span>), the minimum thickness can be estimated:</p>
<p><span class="math-container">$$q_c = \frac{2 E}{[3(1-\nu^2)]^{1/2}} \left(\frac{h}{R}\right)^2 \tag{eq:2}$$</span></p>
<p><span class="math-container">$$ h = \sqrt[3]{\frac{3V}{4\pi}} \sqrt{\frac{q_c}{2 E}\sqrt{3(1-\nu^2)}} $$</span></p>
<p>After having calculated the minimum thickness, it is possible to obtain the total required volume of material <span class="math-container">$V_m$</span> and weight <span class="math-container">$M$</span> by:</p>
<p><span class="math-container">$$V_m = 4\pi R^2 h $$</span>
<span class="math-container">$$M = V_m\cdot \rho = 4\pi R^2 h \cdot \rho $$</span></p>
| 48911 | A query regarding constructing vacuum sealed vessels |
2021-12-26T14:05:02.550 | <p><strong>Background</strong></p>
<p>I have a beginner question concerning rotary vane vacuum pumps like this one: <a href="https://rads.stackoverflow.com/amzn/click/com/B07LBLDZ89" rel="nofollow noreferrer" rel="nofollow noreferrer">VIVOHOME 110V 1/3 HP 4CFM Single Stage Rotary Vane Air Vacuum Pump</a></p>
<p>Here are the pump specifications:</p>
<ul>
<li>Voltage: 110V</li>
<li>Free Air Displacement: 4CFM</li>
<li>Ultimate Vacuum: 5 Pa</li>
<li>Intake Fitting: 1/4 &1/2 Flare</li>
<li>Oil Capacity: 350ml</li>
</ul>
<p>The intended commercial use of this particular pump is for HVAC systems, although I'm not totally sure what the vacuum pump accomplishes in that scenario.</p>
<p><strong>Question</strong></p>
<p>I want to use a pump like this to hold a vacuum in a sealed chamber. Is this possible? Or is a rotary vane pump used exclusively to move fluid/gas?</p>
<p>In particular, I want to build a home cathode ray tube. A vacuum is required so that the electrons don't bump into air molecules as they travel down the tube. I'm not sure if 5 Pa is strong enough vacuum for this application, but the pump is so cost effective that I wanted to look into its operation.</p>
| |fluid-mechanics|pumps|vacuum-pumps| | <p>I have the same pump and I use it to hold (no flow) vacuum without any problems (my use case is to hold vacuum while in epoxy mixture so that all bubbles can be extracted before pouring it into the mold, which usually takes 30min).
But don't expect to get 5 Pa from this pump, I'd say 50 Pa is more realistic.</p>
| 48923 | Can I use a standard HVAC rotary vane vacuum pump to draw a vacuum on a sealed chamber? |
2021-12-26T19:06:34.100 | <p>I don't understand what the hidden lines are nor the fillets tangent to the center line of the circles if someone can help with the 3d isometric that would help a lot. <a href="https://i.stack.imgur.com/z55uV.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z55uV.jpg" alt="I" /></a></p>
| |solidworks|technical-drawing|autocad| | <p>This is just a stiffener that goes around the cylinder that runs across the walls</p>
<p><a href="https://i.stack.imgur.com/pnXTx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pnXTx.png" alt="enter image description here" /></a></p>
<p><strong>Section View</strong></p>
<p><a href="https://i.stack.imgur.com/isQ90.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/isQ90.png" alt="enter image description here" /></a></p>
<p><strong>FUll isometric view</strong></p>
| 48929 | What are these hidden lines tangent to the circle? |
2021-12-26T19:27:53.190 | <p>My assumption is that the taper/conical shape helps separate pure axial/radial loads into a combined load.</p>
<p>For example, a pure axial load should also produce a radial load component in the conical/Timken style bearing.</p>
<p>Is this correct?</p>
| |bearings| | <p>Another interesting feature of tapered roller bearings, is the relationship between axial and radial clearances. Bearings life is usually maximum when there is a slight interference fit -- allowing the rollers to bounce around in a clearance would cause impact loads and thus more wear.</p>
<p><a href="https://i.stack.imgur.com/pAAfc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pAAfc.png" alt="enter image description here" /></a></p>
<p>For some applications, this clearance is adjusted at installation or maintenance. Compared to purely radial bearings, the tapered roller bearings make it easier to make this adjustment, by using a screw of some type to fine-tune the axial distance between the inner and outer races of the bearing.</p>
<p><a href="https://i.stack.imgur.com/p27bB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p27bB.png" alt="enter image description here" /></a></p>
<p>All images from:
<a href="https://www.timken.com/wp-content/uploads/2018/04/Timken-Tapered-Roller-Bearing-Catalog_10481.pdf" rel="nofollow noreferrer">Timken Tapered Roller Bearing Catalog [PDF]</a></p>
| 48931 | Why does the taper on tapered-roller bearings help support axial loads? |
2021-12-27T07:48:55.380 | <p>I read the definition of Isometric Projection from this <a href="https://drive.google.com/file/d/1deyzfegmMlYkxasya7RrP_anBALjnMBg/view?usp=drivesdk" rel="nofollow noreferrer">pdf</a> that Projection plane intersects each
coordinate axis in which the object is defined (principal axes)
at the same distant from the origin.</p>
<p>But we know that projection plane is 2D plane where z-component is either zero or constant, so how is it possible projection plane intersects each coordinate axis?This sentence isn't understanding in that pdf ,what does it mean?</p>
<p>I also read that Projection vector makes equal angles with all of the three principal axes.Isometric projection is obtained by aligning the projection vector with
the cube diagonal.</p>
<p>My question is how to visualize projection from 3D space to projection plane(2D) where diagonal(projection vector) makes equal angle with 3 principal axis of below image(which is projection plane image after projection):
<a href="https://i.stack.imgur.com/lkxz7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lkxz7.jpg" alt="enter image description here" /></a></p>
| |design|technical-drawing|computer-engineering|computer-aided-design|projectiles| | <blockquote>
<p>But we know that projection plane is 2D plane where z-component is either zero or constant, so how is it possible projection plane intersects each coordinate axis?</p>
</blockquote>
<p>The projection plane is like the film / sensor of a camera. It can be in any orientation, not necessarily parallel to the (global) XY plane. So the projection plane can intersect all three axes except in cases where the plane happens to be parallel to (global) XY or (global) YZ or (global) ZX plane.</p>
<p>I used the qualifier "<em>global</em>" for the axes and the planes since one can define another set of XYZ axes where the X and Y axes lie on the projection plane (horizontal and vertical direction). In that frame, Z axis is by definition, perpendicular to the screen / projection plane.</p>
<blockquote>
<p>My question is how to visualize projection from 3D space to projection plane(2D) where diagonal(projection vector) makes equal angle with 3 principal axis ...</p>
</blockquote>
<p>Below are three images which may help you visualize the projection plane, the projection vector / direction and the projected image. The global X, Y ans Z axes are colored red, green and blue. The cube is aligned to the global axes. The projection plane is shown in translucent red square. The projected image of the cube is thick black lines.</p>
<p>The projection direction is shown as thin teal line. Note that all the vertices are extended running parallel to the projection direction until they hit the projection plane.</p>
<ol>
<li>Projection plane is parallel to the XY plane. The cube appears as a square to when projected on the projection plane.</li>
<li>Projection direction lies 45° to the Z and Y axes. Projection plane intersects Z and Y axes. The cube appears to be two rectangles when projected on to the plane.</li>
<li>Projection direction is equal angle from all three axes; i.e., projection direction is the vector <span class="math-container">$[\frac{1}{\sqrt(3)}, \frac{1}{\sqrt(3)}, \frac{1}{\sqrt(3)}]^T$</span> and the angle with each axis is 54.736°. Projection plane intersects all three axis at equal lengths (not visible in the figure attached). Note that, for this projection, the projection line passes through diagonally opposite corners of the cube.</li>
</ol>
<p>1.<a href="https://i.stack.imgur.com/eLfps.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eLfps.gif" alt="projection along Z direction" /></a></p>
<p>2.<a href="https://i.stack.imgur.com/vCf4Z.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vCf4Z.gif" alt="projection along equal angles to Y and Z axes" /></a></p>
<p>3.<a href="https://i.stack.imgur.com/g7ieT.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g7ieT.gif" alt="projection along equal angles to all three axes" /></a></p>
<h2>EDIT</h2>
<blockquote>
<p>I also read that Projection vector makes <strong>equal angles</strong> with all of the three principal axes...</p>
</blockquote>
<blockquote>
<p>Please elaborate how you get <span class="math-container">$[1/(√3),1/(√3),1/(√3)]^T$</span></p>
</blockquote>
<p>Let the projection vector be a unit vector <span class="math-container">$\hat{n} = [a, b, c]^T$</span>. The angle that each axis makes with this vector is given by the dot product.</p>
<p><span class="math-container">$$
\cos\theta = \hat{x} \cdot \hat{n} = [1,0,0]^T \cdot [a,b,c]^T = a\\
\cos \psi = \hat{y} \cdot \hat{n} = [0,1,0]^T \cdot [a,b,c]^T = b\\
\cos \phi = \hat{z} \cdot \hat{n} = [0,0,1]^T \cdot [a,b,c]^T = c\\
a = b = c \quad \text{since} \quad \theta \triangleq \psi \triangleq \phi
$$</span></p>
<p>Since <span class="math-container">$\hat{n}$</span> is a unit vector, <span class="math-container">$a^2+b^2+c^2=1^2$</span>.
So, <span class="math-container">$a = b = c = \frac{1}{\sqrt 3}$</span>.</p>
<p>We can find this vector in another way also;</p>
<blockquote>
<p>aligning the projection vector with the cube diagonal</p>
</blockquote>
<p>The cube diagonal is <span class="math-container">$[1, 1, 1]^T$</span>. This vector has length <span class="math-container">$\sqrt 3$</span>. So the corresponding unit vector is <span class="math-container">$[\frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}, \frac{1}{\sqrt 3}]^T$</span></p>
| 48937 | How to visualize isometric parallel projection? |
2021-12-27T12:07:47.070 | <p>In my course it's preferred not to hatch ribs and webs so should I consider a stiffener like them and not hatch it or is it different and why ? <a href="https://i.stack.imgur.com/tgaKM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tgaKM.png" alt="" /></a></p>
| |solidworks|technical-drawing|autocad|stiffness| | <p>The standard is that The cylinder part should be hatched but not the rib.</p>
<p>So something like the following.</p>
<p><a href="https://i.stack.imgur.com/cB8gE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cB8gE.png" alt="enter image description here" /></a></p>
| 48941 | Should a stiffener be hatched |
2021-12-28T10:15:04.060 | <p>More specifically, if I have an eye bolt being used to lift a body, I can find the various stresses acting on the bolt, but I want to know what happens to the threaded area of the body where the eye bolt is fastened.</p>
<p>Will it be the same thing just in the opposite direction, as in the same shear force? And the area experiencing it would also effectively be the same?</p>
| |fasteners|machine-elements| | <p>When you fix any other face of the body to ground, for example, and then try to apply some forces/moments onto the bolt, then yes the threaded region of the bolt will experience the forces/moments in the opposite direction of the applied loads/moment (this is because the hole's wall is applying equal but opposite load onto the threaded region). While the hole's wall will experience the forces/moments in the same direction as the applied loads/moments. If you consider the bolt and body now as a single unit (since they are glued), then the face of the body which is being fixed to the ground will experience forces/moments opposite to that of applied.</p>
<p>Shear force, or any other force, well it depends how you define the shear force? acting parallel to the faces on threaded region? Well, there might be some forces acting perpendicular to the faces on threaded region as well. But overall, the net force experienced by the threaded region will be the same as the applied load, and in the opposite direction.</p>
| 48948 | Does the nut experience the same forces a bolt experiences but in opposite directions as reaction forces? |
2021-12-28T23:02:04.690 | <p>Most Moody friction factor diagrams (like <a href="https://upload.wikimedia.org/wikipedia/commons/8/80/Moody_diagram.jpg" rel="nofollow noreferrer">this one</a>)</p>
<p><a href="https://i.stack.imgur.com/8ov3T.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8ov3T.jpg" alt="enter image description here" /></a></p>
<p>show a dashed line frequently labelled as either "Complete turbulence", "Wholly turbulent flow" or "Turbulent flow", but I haven't been able to find how to determine this line (and, believe me, I've done some research). One might say that this line is the loci of all the Reynolds numbers for which the friction factor curves assume a flat horizontal shape (given a relative roughness <span class="math-container">$\epsilon/D$</span>), but that doesn't seem to be the case as the dashed line often crosses friction factor curves where those aren't yet flat.</p>
<p>Another curious thing is that I observed that the location of this "Complete turbulence" line seems to disagree from diagram to diagram to a lesser or greater extent.</p>
<p>Any insights on this topic will be appreciated. Thanks.</p>
| |fluid-mechanics|pipelines| | <p>Moody (1944, <em>Trans. ASME</em> <strong>66</strong>(8):671-684) gives the following equation for the onset of complete turbulence:
<span class="math-container">$$\frac{1}{\sqrt{f}} = \frac{\textit{Re}}{200}\frac{\epsilon}{D}$$</span>
and attributes it to a 1943 conference paper by Rouse. I've been unable to obtain a copy of the Rouse paper, but from hints given by Moody, I think the equation represents the surface in <span class="math-container">$\left(\textit{Re},\epsilon/D,f\right)$</span> space on which the laminar sublayer thickness is equal to the surface roughness.</p>
<p>Combining that equation with the Colebrook-White equation (i.e. with the actual observed relationship between Darcy friction factor, Reynolds number, and relative roughness as represented by the solid lines in the Moody chart) gives
<span class="math-container">$$\frac{1}{\sqrt{f}} = -2\log_{10}\left(\frac{57}{\textit{Re}\,\sqrt{f}}\right)$$</span>
i.e.
<span class="math-container">$$\textit{Re} = \frac{57\cdot 10^{1/\left(2\sqrt{f}\right)}}{\sqrt{f}}$$</span></p>
<p>This equation, when plotted, closely matches the version of the dashed line in Moody's own Moody chart. However, it isn't quite such a good match to the dashed line in the Moody chart reproduced (from Wikipedia) in the question (as OP observes, 'the location of this "Complete turbulence" line seems to disagree from diagram to diagram'). Digging through the Wikipedia history of this Moody chart, one discovers one possible explanation: the lines therein were not constructed using the Colebrook-White equation, but using the (slightly less faithful to empirical data, but easier to implement computationally) Swamee-Jain equation. There's nothing wrong with that in itself, except that the Wikipedia Moody chart tries to go to high relative roughnesses and to high Reynolds numbers, beyond the <a href="https://www.chemeurope.com/en/encyclopedia/Swamee-Jain_equation.html" rel="nofollow noreferrer">domain</a> where the Swamee-Jain equation is a good approximation.</p>
| 48956 | How to determine the "complete turbulence" boundary line in a Moody diagram? |
2021-12-29T16:18:11.750 | <p><strong>Update with solution</strong><br />
<em>Solution 1</em><br />
Problem: Beam clamped at left side, free end on right side, point load pointing downwards. x is defined positive from the clamped end towards the free end.</p>
<p>Create FBD from a section x from the free end gives:<br />
<span class="math-container">$M(x) = -P(L-x)$</span><br />
<span class="math-container">$Q(x) = -P$</span><br />
<br />
Differential equation for equilibrium is:<br />
<span class="math-container">$-EI\frac{d\alpha}{dx} = M(x) = -P(L-x)$</span><br />
Integrating this function results in:<br />
<span class="math-container">$EI\alpha(x) = \frac{-Px(2L-x)}{2}+C_{1}$</span><br />
Using the boundary condition that the beam is clamped at <span class="math-container">$x=0$</span>:<br />
<span class="math-container">$\alpha(0) = 0 \rightarrow 0 + C_{1} = 0 \rightarrow C_{1} = 0$</span><br />
Now using the second differential equation for equilibrium:<br />
<span class="math-container">$\frac{d}{dx}\left(\frac{dw}{dx}-\alpha(x)\right)+\frac{p(x)}{\kappa G A} = 0$</span><br />
Integration of this equation leads to:<br />
<span class="math-container">$\frac{dw}{dx} = \frac{Q(x)}{\kappa A G}+\alpha(x)$</span><br />
Substituting in the previous found equation for <span class="math-container">$\alpha(x)$</span> gives:<br />
<span class="math-container">$\frac{dw}{dx} = -\frac{P}{\kappa A G} - \frac{Px(2L-x)}{2EI}$</span><br />
Integrating this once more to find the equation for <span class="math-container">$w(x)$</span> leads to:<br />
<span class="math-container">$w(x) = -\frac{Px}{\kappa A G} - \frac{Px^2(3L-x)}{6EI} + C_{2}$</span><br />
Using the boundary condition <span class="math-container">$w(0) = 0$</span> gives:<br />
<span class="math-container">$w(0) = 0 = -0 - 0 + C_{2} \rightarrow C_{2} = 0$</span><br />
Which gives:<br />
<span class="math-container">$w(x) = -\frac{Px}{\kappa A G} - \frac{P}{EI}\left(-\frac{x^3}{6}+\frac{x^2L}{2}\right)$</span></p>
<p><em>Solution 2</em><br />
Problem: Beam clamped at right side, free end on left side, point load pointing downwards.\</p>
<p>Create FBD from a section x from the free end gives:<br />
<span class="math-container">$M(x) = -Px$</span><br />
<span class="math-container">$Q(x) = -P$</span><br />
<br />
Differential equation for equilibrium is:<br />
<span class="math-container">$-EI\frac{d\alpha}{dx} = M(x) = -Px$</span><br />
Integrating this function results in:<br />
<span class="math-container">$EI\alpha(x) = \frac{Px^2}{2}+C_{1}$</span><br />
Using the boundary condition that the beam is clamped at <span class="math-container">$x=L$</span>:<br />
<span class="math-container">$\alpha(L) = 0 \rightarrow \frac{PL^2}{2} + C_{1} = 0 \rightarrow C_{1} = -\frac{PL^2}{2}$</span><br />
Which gives:<br />
<span class="math-container">$\alpha(x) = \frac{P\left(x^2-L^2\right)}{2EI}$</span><br />
Now using the second differential equation for equilibrium:<br />
<span class="math-container">$\frac{d}{dx}\left(\frac{dw}{dx}-\alpha(x)\right)+\frac{p(x)}{\kappa G A} = 0$</span><br />
Integration of this equation leads to:<br />
<span class="math-container">$\frac{dw}{dx} = \frac{Q(x)}{\kappa A G}+\alpha(x)$</span><br />
Substituting in the previous found equation for <span class="math-container">$\alpha(x)$</span> gives:<br />
<span class="math-container">$\frac{dw}{dx} = -\frac{P}{\kappa A G} - \frac{P\left(x^2-L^2\right)}{2EI}$</span><br />
Integrating this once more to find the equation for <span class="math-container">$w(x)$</span> leads to:<br />
<span class="math-container">$w(x) = -\frac{Px}{\kappa A G} + \frac{P}{2EI}\left(\frac{x^3}{3}-L^2x\right) + C_{2}$</span><br />
Using the boundary condition <span class="math-container">$w(L) = 0$</span> gives:<br />
<span class="math-container">$w(L) = 0 = -\frac{PL}{\kappa A G} + \frac{P}{2EI}\left(\frac{L^3}{3}-L^3\right) + C_{2} \rightarrow C_{2} = \frac{PL}{\kappa A G} + \frac{PL^3}{3EI}$</span><br />
Which gives:<br />
<span class="math-container">$w(x) = \frac{P}{\kappa A G}\left(L-x\right) + \frac{P}{EI}\left(\frac{x^3}{6}-\frac{L^2x}{2}+\frac{L^3}{3}\right)$</span></p>
<p><strong>Note that both solution now produce the same result, but in reverse.</strong></p>
<p><strong>Old message</strong><br />
I am trying to solve the simple problem of a left side supported beam, with a point load at the free end. Lets say the free end is located at L and the force is pointing downwards (negative z-direction). x is defined as positive towards the right (and thus positive towards the free end).
I would arrive at equations:<br />
<br />
<span class="math-container">$Q = -P = \kappa AG(-\phi + \frac{\partial w}{\partial x})$</span><br />
<span class="math-container">$M = -P(L-x) = EI \frac{\partial \phi}{\partial x}$</span><br />
<br />
After which I integrate the second equation, which results in:<br />
<br />
<span class="math-container">$\frac{-P(L-x)}{EI} = \frac{\partial \phi}{\partial x} \rightarrow \phi(x) = \frac{Px(x-2L)}{2EI} + C_{1}$</span><br />
Next I apply the boundary condition: <span class="math-container">$\phi = 0$</span> ax <span class="math-container">$x=0$</span>, which gives:<br />
<br />
<span class="math-container">$\phi(0) = 0 = \frac{P\cdot 0 (0-2L)}{2EI} + C_{1} \rightarrow C_{1} = 0$</span><br />
<br />
Next I integrate the other equation and substitute the found solution and the other boundary condition into it which is <span class="math-container">$w(0) = 0$</span>, which results in:<br />
<br />
<span class="math-container">$-P = \kappa AG(-\phi + \frac{\partial w}{\partial x}) = \kappa AG(-\frac{Px(x-2L)}{2EI} + \frac{\partial w}{\partial x}) \rightarrow \frac{\partial w}{\partial x} = \frac{-P}{\kappa AG} + \frac{Px(x-2L)}{2EI}$</span><br />
<span class="math-container">$\rightarrow w(x) = \frac{Px^{2}(x-3L)}{6EI} - \frac{Px}{\kappa AG} + C_{2}$</span><br />
<span class="math-container">$w(0) = 0 = \frac{P \cdot 0^{2}(0-3L)}{6EI} - \frac{P \cdot 0}{\kappa AG} + C_{2} \rightarrow C_{2} = 0$</span><br />
<br />
However when taking a look at the solution provide on Wikipedia: <a href="https://en.wikipedia.org/wiki/Timoshenko%E2%80%93Ehrenfest_beam_theory#Example:_Cantilever_beam" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Timoshenko%E2%80%93Ehrenfest_beam_theory#Example:_Cantilever_beam</a>, I would expect a solution that provides the same values, except in opposite direction, which is not the case. The difference between my solution and the one provided seems to be created in the first part, which finds the solution for <span class="math-container">$\phi$</span>, but I am unsure what I did wrong. The solution at the extremities is the same, but not the solution throughout the beam itself.</p>
<p>Any help would be highly appreciated.</p>
| |mechanical-engineering|beam|mathematics| | <p>Your moment equation is missing the positive moment, <span class="math-container">$PL$</span> at the left support.</p>
<p><span class="math-container">$$M= PL-P(L-x)$$</span></p>
| 48965 | Solving Timoshenko beam equation for cantilever beam |
2021-12-29T16:49:08.247 | <p>Can anyone just tell me that why doesn't a jumping rope, or any other kind of rope, have any bending stiffness? If I hold one end of the rope in my hand, the rest of the rope will just fall. It won't just keep itself straight. I know it is because of weight of the rope, but why can't a rope carry a bending moment at all? It is very strong in the axial direction, but it can't carry any compression load or any bending loads.</p>
<p>I mean what parameters determines that these structures would be able to resist bending moments but these won't? And why can't it carry compression loads as well? A detailed answer would be extremely appreciated.</p>
| |structures|solid-mechanics|stiffness| | <p>Engineering point of view:</p>
<p>There are ropes that do have significant bending stiffness (and, as a consequence, compression hardness). E.g. steel wire ropes.</p>
<p>This is almost never an advantage - such ropes tend to self-tangle easier (but making intentional knots is harder), tend to break at small bending radius (making most knots unusable) and tend to retain their bent shape after winding/unwinding.</p>
<p>One uses such ropes when the alternatives get impractically thick or impractically expensive.</p>
| 48968 | Why does a jumping rope have no bending stiffness and compression stiffness? |
2021-12-30T11:28:32.637 | <p>Consider a cantilever beam acted upon by a point load at its free end. The shear force diagram is shown.</p>
<p>I have a little doubt,is the shear force at x= 0 ,both 0 and <span class="math-container">$-W$</span> or just <span class="math-container">$-W$</span>? Similarly at the fixed end, is the shear force 0 or <span class="math-container">$-W$</span>, or both?</p>
<p><a href="https://i.stack.imgur.com/DYztlm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DYztlm.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering|stresses| | <p>First of all, the body should always be in equilibrium. This means that the load applied by the support on the body should be equal but in the opposite direction to that what is applied to the body by you. Opposite is important, not the sign or direction. You can assign upwards as positive or downwards as positive, it doesn't matter. If you apply (+) load, then reaction should be equal but (-), but if you apply (-) load, then reaction force should be (+).</p>
<p>It doesn't matter from where you are taking the x, it matters which part of the body (after being cut) are you considering. For example, imagine you cut the part in the middle and x is taken from the left end side where the load is applied. Now, if you are taking the left part of the body, then there needs to exist an internal shear force at the cut location in the opposite direction to the applied load, to ensure equilibrium.</p>
<p>If you take the other part now (which is connected to the support), then the shear force should be in the same direction as the applied load because the only external load acting on it is the reaction load, which is infact in the opposite direction to the applied load. So to make this part become stable, you need to have an internal shear force at the cut location which should be in the same direction and having the same sign as the external applied load, in order to ensure equilibrium.</p>
| 48983 | Shear force in a beam at ends |
2021-12-31T14:56:51.947 | <p>I want to be able to predict the load a single motor and propeller can carry, so I can choose an appropriately sized propeller and powerful enough motor.</p>
<p>I say predict because I'm first going to observe the effects of varying diameter and RPM with a smaller less powerful motor, and once that is complete, I can use the information to get a more powerful motor (and bigger propeller), aiming for thrust force of about 30-50kgf.</p>
<p>I'll be using a tachometer to measure the RPM, a potentiometer to vary the voltage and a (taped down) kitchen scale to measure the thrust force. I plan to power on the motor while on the scale and see how the mass reading decreases, and use that to calculate the force. Is that an appropriate method? Or should I try another way?</p>
| |motors|aerospace-engineering|propulsion| | <p>Unless have really good data for the propeller, motor, and airframe then you cannot "calculate".</p>
<p>Propeller thrust depends on the the RPM, and speed that the propeller is moving through the air. That is a 3-axis graph right there.</p>
<p>Motor speed depends the torque which is a two-axis graph. If the voltage is not fixed then it becomes another 3-axis graph.</p>
<p>And airspeed depends on the speed that the airframe is moving through the air and the drag of the airframe which is a two-axis graph.</p>
<p>So the thrust is determined by an intersection of two 3-axis graphs and one 2-axis graph. It's a big interwined ball of numbers. Not to mention that to calculate any of it from the shape of the propeller or the design of the motor is impractical unless you are a design engineer with really good data and simulation software for those parts.</p>
<p>Your best bet is to use propeller data tables that list the torque and thrust and different combinations of RPM and slipstream speeds.</p>
<p>Then use that with motor torque-speed curves. If you have gearing available then that adds some more degrees of freedom since you now have to adjust the gear ratio so that for each propeller you operate at the motor's maximum power point or maximum efficiency point.</p>
<p>And then estimate the drag of your airframe at various speeds (probably by some rudimentary drag equations and gut feel).</p>
<p>APC Propellers provides data for their propellers under "Technical Data".</p>
<hr />
<p>You did not say how the propeller is being used but you do say "maximize the weight it can support" or "carry weight" which I assume means this is being used to hover like a helicopter. In that case it gets a little bit simpler because your airspeed is zero (since you are hovering) which means your airframe drag is also pretty much zero. So that cuts out 1/3 of your work.</p>
<p>That allows for a less technical approach:</p>
<p>Calculate the no-load RPM of your motor at the desired operating voltage. Most DC motors have peak efficiency of about 70-80% of their no-load speed (about 1/7th of their stall-torque). So hunt around different propeller data tables until you find a good thrust value at zero slip stream speed (since you are just hovering) at your 70-80% of your motor's no-load RPM where the listed propeller input also does not exceed that of the motor. For hovering, you want largest diameter propeller and for a fixed motor power, that will mean the propeller with the lowest pitch. So what you should do is make a list of the lowest pitch propeller available for each diameter. Then run through the data tables for each propeller in order until you find the largest one that won't burn out your motor.</p>
<hr />
<p>You can kind of use rules of thumb relationships, all other things equal (things like diameter, pitch, airspeed, RPM, blade number, and prop geometry. APC blades are basically all the same airfoil within a line: gas, electric, slow flyer), to extrapolate from a few data points like:</p>
<ul>
<li>Motor RPM is proportional to Voltage</li>
<li>Motor torque is proportional to current</li>
<li>Propeller torque is proportional to RPM squared (as per NASA drag equation)</li>
<li>Propeller thrust is proportional to RPM squared (I think, as per NASA lift equation)</li>
<li>Propeller airspeed proportional to RPM</li>
</ul>
<p>Those are the generic well known ones.</p>
<p>There are also a very crudes for one propeller thrust and torque as diameter or pitch vary. I forget them though. I think the only one I looked at was diameter vs torque and diameter vs thrust. I think they were proportional to diameter squared (i.e. prop disc area). You can find them by staring at the APC data with a calculator and trial and error.</p>
<p>I hesitate to combine multiple relations though due to the error. I only ever moved between a single parameter. At least you can ignore varying airspeed.</p>
| 49009 | Predicting thrust of a Motor |
2021-12-31T22:23:15.373 | <p>How many more degrees of tilt can the Millennium Tower in San Francisco withstand before it becomes structurally compromised/unsound?</p>
<p><em>"...An examination in 2016 showed the building had sunk 16 inches (41 cm) with a two-inch (5.1 cm) tilt at the base and an approximate six-inch (15 cm) tilt at the top of the tower.[30] The building is leaning toward the northwest,[30][31][32] and has caused cracks in the building's basement and the pavement surrounding the tower.[33] As of 2018, the sinking had increased to 18 inches (46 cm) with a lean of 14 inches (36 cm).[34]..."</em> Wikipedia, https://en.wikipedia.org/wiki/Millennium_Tower_(San_Francisco)</p>
<p><em>"...The latest data – including the four days that the test pile was installed from Nov. 15 to Nov. 19 – shows a quarter inch of new tilt, as well as a tenth of an inch of settlement at the time the test installation occurred. At the same time, there was marked fluctuation of water pressure below the foundation on the Mission Street side of the structure..."</em> December 7, 2021. <a href="https://www.nbcbayarea.com/investigations/sf-millennium-tower-tilts-quarter-inch-in-four-days/2750189/" rel="nofollow noreferrer">https://www.nbcbayarea.com/investigations/sf-millennium-tower-tilts-quarter-inch-in-four-days/2750189/</a></p>
<p><a href="https://i.stack.imgur.com/WNwiq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WNwiq.jpg" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|structures|solid-mechanics| | <p>Modern building failure due to tipping is rare but did occur, usually under strong earthquakes and loss of foundation (soil and structure), as the tilt will shift the gravity center of the building that induce a tremendous amount of stress on the foundation piling and the soil mass surrounding the piling.</p>
<p>So, how much is too much for a building to handle (remain functional)? As the building code generally does not specifys limit on inclination other than service considerations (floor levelness, shaft plumbness...), the answer can only be provided by the structural designer who knows the building structure, the soil strength, and has a handle on soil-structure interaction.</p>
<p>But for a rought estimate, H/400 - H/500 are the most recommended limit on "drift" due to wind, deflection beyond such limit is believed to cause severe structural, and/or foundation damages. From that, you can do your math by assuming H/500 is the absolute maximum a building can tilt without failure (Note, the safety factor for wind load is quite large, that can approaching or beyond 2).</p>
<p>Note, another building failure mode is settlement, however, it creates more of service issues and localized structural damages, especially in the case of uniform settlement. So it is not addressed here.</p>
<p><a href="https://i.stack.imgur.com/IRvsm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IRvsm.png" alt="enter image description here" /></a></p>
| 49021 | How many more degrees of tilt can the Millennium Tower in San Francisco withstand before it becomes structurally unsound? |
2021-12-31T23:46:03.420 | <p>I drew the outer hollowed body and the cylinder on 2 separate parts but I can't figure out how to draw this rib so that it is passes through the centre of the cylinder, a video of a similar example would help a lot<a href="https://i.stack.imgur.com/p3X7u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/p3X7u.png" alt="" /></a></p>
<p><a href="https://i.stack.imgur.com/fbOmP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fbOmP.jpg" alt="enter image description here" /></a></p>
<p>Here is my work so far <a href="https://i.stack.imgur.com/iVwu2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iVwu2.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/kiNmJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kiNmJ.png" alt="enter image description here" /></a></p>
| |modeling|cad|technical-drawing|autodesk-inventor| | <ol>
<li>It looks like the part is symmetrical about the YZ plane.</li>
<li>Create a new sketch on the YZ plane and draw the rib. Extrude it symmetrically. You may need to work in wireframe or "show hidden edges" (or whatever SolidWorks calls that).</li>
<li>No, you wouldn't create a separate part. The piece is all one.</li>
<li>I have no idea why you posted a picture of a cylinder.</li>
</ol>
<p><a href="https://i.stack.imgur.com/DMPcQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DMPcQ.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. One way of creating the part is to draw half of it and mirror when complete. Screengrab from <a href="https://onshape.com" rel="nofollow noreferrer">OnShape</a> model.</em></p>
| 49023 | How to draw hidden rib in Autodesk Inventor |
2022-01-01T03:21:45.987 | <p>I'd want to build a scale which can be "locked" in order to prevent weight from being pressed in it.</p>
<p>The scale and the weight are permanently fixed on a vehicle and
I want to make sure it doesn't break under excessive force during travel.</p>
<p>What I have in mind is something like the schema below. Does it have a name?</p>
<p>Or are there other, more common ways to achieve this? I am looking.at something that can be done quickly (a screw would take too long and I'd need 4 of them.</p>
<p>If it matters, I need about a 5mm displacement, and a supported weight of 200kg</p>
<p><a href="https://i.stack.imgur.com/P4oxc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P4oxc.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering| | <p>Maybe I'm not understanding what you have in mind but your drawing looks really complicated with sliding rollers in captive grooves and such. I also imagine the lateral movement when weighing something is undesired. I would just try to shove something underneath.</p>
<p>Simple methods are vertical guide rails that can be pinned. Or a swinging beam with a radiused top or eccentric cam that can be flipped up and locked/pinned into place.</p>
<hr />
<p>But in my opinion, your thought about screws is the best. By far the most secure and with the greatest leverage. Use one in the form of a screw jack and use a multi-start thread to make it faster. That will solve the speed issue. You'll need to find someone with a lathe to single-point thread you some, which means instead of screws you might as well get them to just make two big screw jacks with hand crank wheels, instead of four smaller, regular screws.</p>
<p>Or if you can't find someone with a lathe, then perhaps find
<a href="https://www.mcmaster.com/multiple-start-rods/" rel="nofollow noreferrer">multi-start/fast travel threaded rod or lead screws</a>. Typically used for linear motion.</p>
<p>If your plate was being guided on vertical tight rails then you only need one screw jack if it is big and strong enough since the rails would eliminate tilting.</p>
| 49027 | What is the name of this height-adjustment structure? |
2022-01-01T04:21:38.343 | <p>Solid Mesh vs Surface Mesh, which method should I choose in simulation studies. What is the geometry behind these meshes. Which mesh gives accurate results?</p>
| |mechanical-engineering|solid-mechanics|simulation|ansys|meshing| | <p>Both of them <em><strong>potentially</strong></em> give accurate results. The accuracy of the results depends mainly on the application and the correct use.</p>
<p>As a general rule</p>
<ul>
<li>Surface meshes are usually more appropriate with sheets of metals. (the width and length have values that are much greater than the thickness).</li>
<li>Solid meshes are usually more appropriate for parts that all the dimensions are in the same order of magnitude.</li>
</ul>
<p>Apart from that basic rule, there are many asterisks.</p>
| 49028 | What is meant by Solid Mesh and Surface Mesh? |
2022-01-01T09:21:55.883 | <p>As a high school student, I am trying to learn more about airfoils and the theory behind it.</p>
<p>I started off with analyzing the NACA0012 airfoil on Autodesk CFD. I modified the lower surface as shown in the image below (the modified airfoil simulation at Angle of Attack=9 degrees).</p>
<p>I simulated the NACA0012 airfoil and the modified airfoil at Reynolds number 1 million (with inlet velocity of 48.655 m/s, chord length of 0.304 8m (1 foot), air density=1.20473 kg/m<sup>3</sup>, coeff of viscosity= 1.817×10<sup>-5</sup> Pa). My simulations were incompressible flow simulations.
<a href="https://i.stack.imgur.com/7P7F7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7P7F7.png" alt="The modified airfoil at AoA=9" /></a>
I know that the NACA 0012 airfoil has its center of pressure at quarter chord point at low angles of attack by the thin airfoil theory.</p>
<p>In the case of my modified airfoil, I find that the centre of pressure moves from 0.596 chord length to 0.259 chord length as I vary the angle of attack from 0 to 10. I have attached an image of the center of pressure location below. The baseline airfoil is the NACA0012 airfoil and I have measured location of center of pressure from the trailing edge.</p>
<p><a href="https://i.stack.imgur.com/EaeKG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EaeKG.png" alt="enter image description here" /></a></p>
<p>I also read that beyond the viscous fluid layer, near a solid, the fluid behaves like a frictionless fluid. Based on this theory, I assume that at low angles of attack, the fluid slows down and pressure increases by Bernoulli's principle beyond the viscous layer as shown in the image below.</p>
<p><a href="https://i.stack.imgur.com/ZNTGm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZNTGm.png" alt="enter image description here" /></a></p>
<p>I think this idea reasonably explains why the centre of pressure is closer to the trailing edge:the high pressure fluid exerts more force near the trailing edge as compared to the NACA0012 airfoil, which moves the center of pressure toward the trailing edge.</p>
<p><a href="https://i.stack.imgur.com/MTBy9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MTBy9.png" alt="enter image description here" /></a></p>
<p>The NACA 0012 airfoil at AOA 0 for comparison.</p>
<p>Based on the viscous boundary layer theory or any other theory that I am unaware of, is it possible to explain the reason for the centre of pressure moving toward the quarter chord point at higher angles of attack?
I am thinking that the viscous layer is thicker at higher angles of attack and hence the freestream fluid velocity does not decrease as much as it does at angle of attack zero. I am pretty sure this is incorrect and there is a lot more going on. Would someone mind explaining what other factors are at play or what other topics I need to know to answer this question?</p>
<p>Appreciate any help.</p>
| |fluid-mechanics|airfoils| | <p>Theories are developed to represent something happening in reality, in a form of mathematical model. Theory is not the benchmark, reality is. You cannot use a theory to prove something which is happening in reality, but can only develop a theory so that this reality can be represented correctly. Sometimes, using a specific theory to gain reasoning for a different real occurence can lead your mind getting confused and not reaching any satisfying conclusion.</p>
<p>I don't know much about boundary layer and its relation to the movement of center of pressure with changing angles of attack, but I know this that as soon as you start increasing the angle of attack for any type of airfoil, then most of the pressure forces and shear forces are experienced at the airfoil's front (near the leading edge) rather than at the back (trailing edge), as proven by extensive experiements. This ofcourse has to do with the changing airflow around the airfoil, when increasing the angle of attack. Look at the figure below (taken from <a href="http://avstop.com/ac/flighttrainghandbook/pressuredistribution.html" rel="nofollow noreferrer">this link</a>).</p>
<p><a href="https://i.stack.imgur.com/J6LTc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J6LTc.png" alt="enter image description here" /></a></p>
<p>This asserts that the center of pressure should move forward because most of the amount of forces are now acting forward, and by definition, center of pressure is the location where the net forces on the complete airfoil act.</p>
| 49034 | Why does centre of pressure move forward with increasing angle of attack in this airfoil? |
2022-01-01T12:30:28.387 | <p>It's been kind of painful to create a rib I'm new to inventor while I was trying to create a rib symmetric about the YZ Plane this error keeps popping up<a href="https://i.stack.imgur.com/NsUe3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NsUe3.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/LhEoG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LhEoG.png" alt="enter image description here" /></a></p>
| |modeling|cad|technical-drawing|autodesk-inventor| | <p>As explained in my answer to your <a href="https://engineering.stackexchange.com/questions/49023/how-to-draw-hidden-rib-in-autodesk-inventor/49036?noredirect=1#comment90482_49036">previous question</a> you just need to create a closed loop sketch on that plane and extrude it 5 mm (or 10 mm symmetrically). There's no need for use of the rib tool.</p>
| 49038 | Autodesk inventor create rib feature failed |
2022-01-01T15:07:08.273 | <p>So I used AutoDesk inventor and created this machine part for the purpose of exercise , But it is preferred to not to hatch ribs in my course so is there a way to make it look like image (5 and 6) ?<a href="https://i.stack.imgur.com/2kuul.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2kuul.png" alt="enter image description here" /></a></p>
<p>original
<a href="https://i.stack.imgur.com/AdmSZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AdmSZ.jpg" alt="enter image description here" /></a></p>
<p>3D Views<a href="https://i.stack.imgur.com/ZVQaU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZVQaU.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/mHcQ6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mHcQ6.png" alt="enter image description here" /></a></p>
<p>image 5
<a href="https://i.stack.imgur.com/bujnR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bujnR.png" alt="enter image description here" /></a></p>
<p>Free hand sketch of what it should look like<a href="https://i.stack.imgur.com/tFanu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tFanu.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|modeling|cad|technical-drawing|autodesk-inventor| | <p>This is the jog that John Holtz referred to I believe. I did the following in Inventor LT.</p>
<p><a href="https://i.stack.imgur.com/1mVaX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1mVaX.png" alt="enter image description here" /></a></p>
<p>What I noticed and not sure if there is an option for this is the vertical line in the bottom plate where the section line jogs. Personally I do not believe there should be a line here as there is no change in material type or dimensions at the jog, but I am not an expert in the field.</p>
<p>UPDATE</p>
<p><a href="https://i.stack.imgur.com/2yiK3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2yiK3.png" alt="enter image description here" /></a></p>
<p>The above was done in Inventor. I had to jog the section line in order to get the geometry. I then had to add a "sketch" to the drawing. This is where I clicked on individual lines and turned their visibility off (section jog lines). I also had to turn off the automatic hatch in Section F-F.</p>
<p>I then drew in the boundary lines for the rib. I then projected the required section lines into the sketch. Once all the lines were in place I did a manual hatch of the area.</p>
<p>I did a similar thing for section E-E. I had to project the right side line work and manually hatch. The hatch scale did not match so I had to adjust the hatch scale so it would match.</p>
| 49041 | Are these section views correct Autodesk inventor |
2022-01-02T23:33:29.787 | <p>I'm trying to make a simple design that consists of two ball joints connected to each other in such a way that I have rotation and translation of the second joint, but with an adjustable resistance. I'm no engineer, so I'm unsure about the whole thing, and I might end up using different types of mechanical joints altogether. How can I add resistance to a ball joint? Is there any well-known two joint structure that resembles what I want?</p>
| |mechanical-engineering| | <p>The ball joint can be seated in a split-spherical shell socket that is tightened by screw. The one below uses a hydraulic mechanism to tighten all the clamps present with a single knob, but other models use a wire. Of course a direct bolt could also be used per joint.</p>
<p>Most obviously seen in the photo on the "wrist joint" and the "shoulder joint". It's dark, but look closely, you can see the split in the spherical shell that allows how tightly it clamps the ball to be adjusted.</p>
<p>Actually thinking about it, it might actually draw the ball into a wedge shaped groove to exert clamping pressure rather than closing the spherical half-shell around the ball.</p>
<p><a href="https://i.stack.imgur.com/RjdU7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RjdU7.jpg" alt="enter image description here" /></a>
Mitutoyo 7033B</p>
| 49055 | How do I add resistance to a ball joint? |
2022-01-03T02:55:21.140 | <p>I am working on building an autonomous snow thrower similar to the one in this <a href="https://youtu.be/1NN8v8WS44c?t=42" rel="nofollow noreferrer">video</a>.</p>
<p>I tore an old <a href="https://www.tractordata.com/ltphotos/F001/1924/1924-td4-b01.jpg" rel="nofollow noreferrer">RX73 tractor</a> down to just the chassis, by removing the engine, transmission, mower deck, brakes ... basically everything. Here are pictures of the <a href="https://i.stack.imgur.com/u4PVX.jpg" rel="nofollow noreferrer">Top of Frame</a>, and the <a href="https://i.stack.imgur.com/NgsOb.jpg" rel="nofollow noreferrer">Underside of Frame</a> after I cleaned it a bit.</p>
<p>In terms of steering and control, <a href="https://en.wikipedia.org/wiki/Differential_wheeled_robot" rel="nofollow noreferrer">differential drive robots</a> are dead simple to control vs. ackerman steering (traditional car steering).</p>
<p>It's often preferred in robotics to use two motors, one connected to each back wheel, and use a free caster wheel in the front, or use a chain to connect each back wheel to the front wheel on its same side.</p>
<p>Another popular method is all-wheel drive by connecting a motor to each wheel but use the same DC output for the front and back motors on each side.</p>
<p>Seeing as I already just <strong>two</strong> 250W Motors, and chains, it seems like the easier option is to use a chain to connect the front and back wheels on each side, rather than buying more motors.</p>
<p>I am concerned however, with this design, that the chain in exposed to the ground seeing as the underside of the chassis is not enclosed.</p>
<p>I image this could potentially cause snow to build up somewhere in that mechanism. If you look at the video and how the chassis is connected to the thrower, there is a pair of linear actuators with arms, so I can also switch out the thrower for the mower deck and this robot to mow the lawn next summer. So I am concerned about dirt, mud and grass getting stuck as well.</p>
<p>My questions are:</p>
<p>Are my concerns valid, or just lack of experience? If valid whats a good way to prevent that from happening?</p>
<p>What would you change about the drive mechanism? An additional two motors is not out of the question, they're approx $40 each</p>
| |mechanical-engineering|structural-engineering| | <p>Let's first think about the snow blower function.</p>
<p>Depending on where you live and if there is salt (may be by road cleaning trucks ) in the environment the exposed chains' grease will shave and collect and salty dust and particles from pavement and form a sticky paste, making it hard to keep them from rusting. The chains tend to throw this stuff to under carriage too and cause damage.</p>
<p>As for attaching a mower, again an exposed chain is bound to get jammed in small surface roots and old dead plant sticks.</p>
<p>you could pick heavy duty chains and deal with occasional cleaning and maintenance, but your idea of having a robot control means you don't want to constantly observe the machine.</p>
<p>So your idea of using separate motors for front wheels make more sense.</p>
| 49060 | Concerned about snow and dirt getting into the drive chain based on my design |
2022-01-03T10:20:30.970 | <p>I've attached a snap of the part I'm unable to understand <em>Machinery's Handbook 29th Edition</em>.
Here's a link to the ebook as well <a href="https://www.vtc1.org/cms/lib/PA03000913/Centricity/Domain/21/Machinerys%20Handbook%2029th%20Edition.pdf" rel="nofollow noreferrer">https://www.vtc1.org/cms/lib/PA03000913/Centricity/Domain/21/Machinerys%20Handbook%2029th%20Edition.pdf</a><br />
(page 1537)</p>
<p><a href="https://i.stack.imgur.com/OG34N.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OG34N.png" alt="enter image description here" /></a></p>
<p>I'm struggling to understand the relevance of this value Q that I arrive on. It gives me the minimum thread engagement length but doesn't factor in the load on the internal/external material.</p>
<p>So I get the thread engagement length from just the material properties.</p>
<p>What if I had the same materials and the bolt was only bearing a 0.5 N load, and my engagement was only 5 mm. Will the threads still strip? I think this recommended length has some other significance. But I need someone's help in understand that.</p>
| |machine-design|fasteners|threads|machine-elements| | <p>The minimum length of engagement ist the length of engagement that (computationally) will ensure that overtightening the screw will snap the screw, and not strip the threads.</p>
<p>The external and internal threads (that are engaged) need to carry more load than the shaft, for this to happen, so both external and internal thread material properties come into play.</p>
| 49065 | Unable to understand minimum length of engagement for bolts |
2022-01-04T13:29:06.287 | <p>I wish I could directly embed this video here</p>
<p><a href="https://youtu.be/Syb3MHtjojQ" rel="nofollow noreferrer">https://youtu.be/Syb3MHtjojQ</a></p>
<p>It can be seen that no external chemical has be put into the bottle, so the bottle content has clearly not been modified.</p>
<p>A friend of mine said it must be capillary action, but it doesn't look like it one to me.</p>
<p>It's doesn't look normal CO<sub>2</sub> release either, since it flows steadily for 40+ seconds.</p>
<p>Assuming no battery/pump has been attached to the straw, how can this be explained scientifically?</p>
| |experimental-physics| | <p>It is CO<sub>2</sub> that is released from the liquid. As the liquid warms up, it is unable to hold as much of the CO<sub>2</sub> in a dissolved state. The CO<sub>2</sub> raises the gas pressure above the liquid. I think that you can figure out the rest.</p>
| 49088 | Dissolved gas release from liquid |
2022-01-04T14:48:00.257 | <p>It's always said that jerk function should be smooth to avoid undesired vibrations. What exactly is the vibration in a cam follower mechanism and why is it important to avoid it?</p>
| |mechanical-engineering|mechanisms|solid-mechanics|machine-design|machine-elements| | <p>Response of the follower to a cam curve command creates two types of vibrations.</p>
<ul>
<li><p>A steady-state vibration.</p>
</li>
<li><p>A transient type of vibration.</p>
</li>
</ul>
<p>These vibrations should be designed not to be close to the natural frequency of the camshaft or connecting mechanism.</p>
<p><a href="https://pure.tue.nl/ws/files/3734499/169394.pdf" rel="nofollow noreferrer">source</a></p>
<p><a href="https://i.stack.imgur.com/2nc3V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2nc3V.png" alt="cam jerk" /></a></p>
| 49089 | Question about cam follower design |
2022-01-05T13:33:11.200 | <p>I am replacing the ceiling in my flat with some soundproofing materials and since these are heavier than usual materials used I want to check if the new materials would pose a problem.</p>
<p>The room I want to apply this to is 3.3x4 metres.</p>
<p>Joists
Height: 22.2 cm / 8.85 inch;
Width: 5 cm / 2 inch;
Length: 330 cm / 130 inch</p>
<p>30 cm spacing between these;
13-14 joists</p>
<p>What I want to know is how I can calculate the maximum load that the ceiling could hold.</p>
<p>Any ideas?</p>
| |structural-engineering|structural-analysis| | <p>You need to know what type of lumber is used in your ceiling to know its strength for moment and shear,<span class="math-container">$F_b \ , V_{allowable}$</span> and its elastic modulus, E. but for illustration assume a lower range of</p>
<p><span class="math-container">$$F_b=1000psi \ and\ V=100psi$$</span></p>
<p>Then your moment is</p>
<p><span class="math-container">$$M=\omega l^2/8= \omega*(11*12)^2/8=2178\omega$$</span></p>
<p>you beam elastic module <span class="math-container">$S_x$</span> is
<span class="math-container">$$S_x=bh^2/6=2*8.8^2/6=25.80in^3$$</span></p>
<p><span class="math-container">$$\sigma x=F_b=1000=M/S_x=2178\omega/25.8=84.4\omega$$</span>
<span class="math-container">$$\omega= 1000/84.4=11.8lbs/ft^2$$</span></p>
<p>This is the allowable load per square foot your ceiling joist can support.</p>
<p>Usually, Fb is in the range 1500-1800psi for Douglas fir and similar structural lumber, implying your ceiling can support approximately 16-17lbs.ft^2. then we have to check for deflection and shear as well. Deflection <span class="math-container">$\delta$</span> should be smaller than l/360</p>
<p><span class="math-container">$$\delta=\frac{5 \omega l^4}{384EI}$$</span></p>
| 49115 | How much load could the ceiling hold |
2022-01-05T14:32:57.737 | <p><strong>Summary:</strong> Are there any worthwhile methods for rough calibration of CO2 sensors without proper chemistry lab equipment, calibration gas, or other known-good sensor available? It would only be used as an air quality indicator.</p>
<p><strong>Full story:</strong> I have a couple of <a href="https://media.digikey.com/pdf/Data%20Sheets/DFRobot%20PDFs/SEN0159_Web.pdf" rel="nofollow noreferrer">SEN0159</a> analog CO2 modules based on the <a href="https://eu.mouser.com/datasheet/2/830/CO2b_MG811_datasheet-2488848.pdf" rel="nofollow noreferrer">MG-811</a> sensor. According to the manufacturer, calibration has to be done manually for each individual sensor, i.e. to determine the parameters for the conversion function from voltage to concentration percentage. There are two sets of parameters mentioned as examples in the documentation, but these differ wildly from one another, so indeed these particular values cannot be trusted.</p>
<p>For the calibration, one would need to expose the sensor to at least two samples of known CO2 concentration. One of these would obviously be the baseline value, in fresh outdoor air, which can be treated as 400 ppm (or nearby meteorological station data could be consulted). For other type of sensors, a zero ppm calibration point would be available to use with e.g. pure nitrogen gas. This particular sensor has a measurement range of 400 to 10000 ppm though, meaning that some other point has to be used within those bounds.</p>
<p>So the challenge is to obtain an approximately known CO2 concentration in some container (with the sensor in it). I am asking in Engineering with the hope to get some creative answers using common workshop/household tools, cheap material, and widely available CO2 sources like soda siphon, carbonated water, burning something etc, maybe even simple exhaled air.</p>
| |measurements|sensors| | <p>If you're really only looking for a rough calibration, you may be able to set up a dilution system. 10,000 ppm is 1%, so you'd be aiming for a gas mixture of 1% CO2 and 99% air. Use an air compressor and small flowmeter to get the air at some known flowrate, say 100 ml/minute. Then put some dry ice in a mostly-sealed box. As it sublimates it will displace the air and eventually be filled with pure CO2. Now you need 1 ml/min drawn out of the box and combined with the air flow. You'll need a little pressure to drive it, so you'd need a small pump to draw the CO2 out and push it through the flowmeter into the combined stream.</p>
<p>Acquiring all those bits and pieces may well cost more than buying or renting a calibrated reference sensor.</p>
| 49116 | DIY CO2 sensor calibration |
2022-01-05T20:09:22.327 | <p>I have a question regarding a system containing a water source, with two pipes connected using a T. There are two different states of the system; one being where both pipes are open and water is flowing, the other being that one pipe is closed.</p>
<p>Is it possible to have a constant flow out of each pipe, in both situations? If so, what device is needed to keep the constant flow?</p>
<p>Example:</p>
<p>Pipe 1 and 2 is open. 15 L/min is flowing out of both pipes. Pipe 1 is closed, but 15 L/min is still flowing from pipe 2. Image related.</p>
<p><img src="https://i.stack.imgur.com/gvQ5h.png" alt="Magical_device" /></p>
<p>The two things I have come up with are:</p>
<ul>
<li>Flow control valves at the start of each pipe, ensuring no flowrate higher than what is set</li>
<li>A pressure reducing valve, ensuring that the pressure is constant on the valve-side.</li>
</ul>
<p>This has led me to two things; firstly that adding a flow control valve would be costly when expanding the system, and secondly that I do not know enough about fluid dynamics to understand how a pressure reducing valve could work in relation to mass flow rate.</p>
<p>Thanks for the help.</p>
| |fluid-mechanics| | <p>First of all, having the same flow in both pipes without flow control valves on both is set by the characteristics of the dwnstream piping in each. Unless they are mirrors of each other, all valves open will likely not have the same flow in both. You could throttle one if desired to make them the same if that is a goal.</p>
<p>Second, you could control flow through one or the other when one is shut off via a throttle with specific settings (your "device to control flow"). You could determine the settings via experimenting, the setting might be different with Pipe 1 closed versus 2. This assumes that the valves for Pipes 1 & 2 are either open or shut.</p>
<p>If you want a specific flow rate, then flow controllers on each pipe with control valves is really the only way to go.</p>
<p>Also, I'm not sure what the "pressure reservoir" is for, water doesn't really need this.</p>
| 49121 | Constant flow rate from two closable pipes with single source |
2022-01-06T09:30:07.890 | <p>All wind turibne use low lift coefficent airfoils from 1.0 to 1.5.</p>
<p>Why not use high CL airfoils to get maximum torque?</p>
<p><a href="http://airfoiltools.com/airfoil/details?airfoil=s812-nr" rel="nofollow noreferrer">http://airfoiltools.com/airfoil/details?airfoil=s812-nr</a></p>
| |aerospace-engineering|aerodynamics|wind-power| | <p>I've finally got a few minutes to try to answer this.</p>
<p>Torque is bad. It's necessary, but it is bad. Ideally, you want the turbine to spin infinitely fast at almost zero torque, not slowly at high torque.</p>
<p>To understand why, lets look at how energy is extracted from the air and transferred to the rotor. If we take an earth reference frame aligned to the free stream air, and have a rotor plane that is perpendicular to that freestream, then the freestream has a velocity components of u=wind speed, v=0, w=0. Downstream of the rotor, we have u=ud, v=vd, w=wd. sqrt(vd^2 +wd^2) is the wake swirl velocity in the rotor plane. Wind speed - ud is the change in axial velocity of the flow.</p>
<p>The transition between the freestream conditions and the wake conditions happens gradually. The swirl acceleration and axial flow deceleration begin well upstream of the rotor and continue well downstream of the rotor. It turns out to be rigorously true that the conditions at the rotor are halfway between the freestream and the fully developed wake. We adjust the freestream flow to the conditions at the rotor with an axial induction factor and a swirl induction factor. Both vary across the radius of the rotor. Then we account for the blade motion to compute the actual angle of incidence of the wind to the spinning blade in the rotor plane.</p>
<p>So now we have a new reference frame in which the blade element is fixed, and the speed and angle of the incident air is correctly accounted for. Since the <em>blade element is fixed</em> in this reference frame, the air does no work, and energy and momentum are conserved as the flow develops from freestream to fully developed wake. This is only true in this blade element frame of reference.</p>
<p>To get maximum power out, we need to maximize the product of mass flow through the rotor <em>times</em> the energy change per unit mass of air. As the axial velocity slows down, the air stream has to get fatter to preserve mass continuity, so air goes around the rotor instead of through it and mass flow is reduced. If you don't slow the air down, all the air goes through the rotor, but you don't get any energy out of it, so power is zero. If you slow the air way down, the energy extraction per unit mass is high, but the mass flow is very small, so power is very low. You need to create a modest axial velocity change which permits a high mass flow rate to get best performance.</p>
<p>So what has this to do with torque? Driving torque creates an equal and opposite reaction in the wake - wake swirl. <em>All</em> the swirl energy in the wake is energy that is being robbed from the rotor. The only way the wake can gain swirl velocity is at the expense of axial velocity (conservation of energy). So swirl causes a lower downstream axial velocity, and that lowers the mass flow rate through the rotor. This is the hard part, that the force from the fixed rotor causes a swirl momentum, but when you apply conservation of energy, the axial velocity has to slow down to keep the system energy balanced. This is not at all intuitive, because there doesn't appear to be any force slowing the air axially. But there is a force, and it is called induced drag.</p>
<p>So at this point, there is an ideal wake shape that can be computed. It has some swirl and some radial expansion at the rotor plane, and the axial flow is somewhat slower than free stream. But the shape, ie the relative size of the induction factors, is the same for all HAWTs under ideal conditions. And when you design a blade set to produce this ideal wake (called inverse design) and account for friction and structural mechanics, you don't want a high Cl blade. You want high rpm (high tip speed ratio), low blade count, low solidity, and low torque.</p>
<p>This entire argument is usually handled a bit differently from a mathematical standpoint because the relationship between axial velocity and swirl velocity can be gotten at in different ways. But wake swirl lowers the mass flow through the rotor and lowers the available energy per unit mass as well. So you want to make it small. And wake swirl is torque.</p>
| 49129 | Why wind turbine dont use high lift coefficient airfoils? |
2022-01-06T13:37:31.003 | <p>I’m looking for a way to discern whether a DAB radio plugged into a non-smart electrical outlet is switched on or off. Note that the radio is constantly plugged into a live socket, there is a power button on the device that turns it "on" and "off".</p>
<p>I'm not sure how much the power draw changes when the device is on as opposed to in standby but I'm not sure it would be much as it's just a low-power radio.</p>
<p>Is there any way for a device like a Raspberry Pi to tell whether the radio is on?</p>
| |electrical-engineering| | <p>I had a similar requirement wanting to know when my oil-filled electric heater was switched on so I could remember to turn it off before going to bed. The heater doesn't have a nice red LED to show that its working, nor a simple on/off switch. Going back to basics I thought that I could sense the current flowing through the mains cable. If you are happy working with mains electric and take adequate safety precautions you can intercept the mains lead into an enclosure and separate the two insulated conductor wires. Then you could detect the magnetic field generated by the current passing through one wire. You do not have to expose the copper conductor just make sure that the two insulated wires are apart from each other enough so the magnetic fields generated by the alternating current don't cancel each other out. The magnetic field can be detected by a coil wound round one of the conducting wires. A voltage will be induced in the coil which can be detected by the ADC (analog to digital converter) on the pi. Because of the electronic on/off switching there may be a detectable magnetic field even when the radio itself is not switched on but I would expect you would be able to detect a significant change when the radio is working. This idea is just a sketch which I am currently working on but if its appealing then its a direction you could go in. You could also use a Hall Effect device. Wiser members may expand on this.</p>
| 49130 | Is there any way to sense whether a mains DAB radio is currently turned on? |
2022-01-08T00:09:25.303 | <p>Say you have a very complex <strong>completely mechanical machine</strong>, and you want to put a dial indicating rotation <strong>speed</strong> of the central drive shaft and its <strong>torque</strong>. But the machine is <strong>completely mechanical</strong>, and you <strong>don't have pneumatics or hydraulics</strong> in it. Is it possible to create a speed dial or a torque dial simply using springs and different types of gears? I would imagine an internal clock of sorts would be necessary for the speed dial, and that could be done. Any help would be much appreciated</p>
| |mechanical-engineering|gears|mechanisms| | <p>to read torque mechanically, you mount the engine or motor spinning the shaft on a hinge along one side and a compression spring along the other. when the engine is applying torque to the shaft, the hinge allows the motor to rock slightly, compressing the spring, and a needle attached to the motor will then read torque on a stationary scale next to it.</p>
<p>BTW note that an ordinary automobile speedometer contains gears and a rotating magnet near an aluminum disc. That magnet rotates via a flex shaft that runs to one of the front wheels of the vehicle. When the magnet rotates it applies a torque via eddy current induction to the disc in proportion to its speed. a small hairspring preloads the disc, to which the speedo needle is attached, so it reads zero when the magnet is not moving. it hence reads RPM without electronics or software.</p>
| 49150 | Is it possible to construct geared speed and torque dials? |
2022-01-08T05:03:52.517 | <p>How to create a schematic that shows pressure and force distribution around a blunt body (for e.g. a cylinder) and an airfoil? Where can one get the data for these distributions? I looked for these schematics in some of the well known books on aerodynamics (By John Anderson for e.g.) I couldn't find them. Referring me to a literature that has these schematics would be helpful.</p>
| |fluid-mechanics|airflow| | <p>Ansys has versatile CFD applications, and many free sites with free examples and some free books, some discounted books. It offers student prices and some start-up versions with lower prices.</p>
<p>one of the sites that have some free tutorials is this <a href="https://cfd.ninja/ansys-fluent/ansys-fluent-drag-coefficient-reference-values-cube-3d/" rel="nofollow noreferrer">Ansys</a></p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/lbOnA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lbOnA.png" alt="a cube drag." /></a></p>
| 49159 | Pressure and friction force distribution around a blunt body and a streamlined body |
2022-01-08T11:42:57.890 | <p>The book I'm referring to for studying beams, said that its quite cumbersome to construct shear force and bending moment diagrams, by writing the SF and BM equations for each segment of the beam and then plotting the equations. To solve this problem we can form relationships between shear force, Bending moment and loads and then use them to construct SFD and BMD.</p>
<p>So, the book starts off by developing a relationship between shear force and load. In doing so, it says that consider a beam with some arbitrary distributed load. At any distance x, we take an element between two sections. The distributed load on this element can be assumed uniform with intensity say, <span class="math-container">$q$</span>.</p>
<p><a href="https://i.stack.imgur.com/IBYuj.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IBYuj.jpg" alt="enter image description here" /></a></p>
<p>Then it says that shear force and bending moments will be developed at each section of this element, and in general they can vary along the length. <strong>The book then assumes the direction of shear force and BM to be what it considers "positive"</strong>.</p>
<p><em>The author has taken shear force as positive if it tries to rotate the element clockwise, and BM positive when it tries to compress the upper part of the element.</em></p>
<p>applying the equilibrium condition to the element yields,</p>
<p><span class="math-container">$$V_1 - V_2-qdx=0$$</span>
<span class="math-container">$$V_2-V_1=-qdx$$</span>
<span class="math-container">$$\frac{V_2-V_1}{dx}=-q$$</span></p>
<p>Thus <span class="math-container">$V_2-V_1$</span> will be differential</p>
<p><span class="math-container">$$\frac{dV}{dx}=-q$$</span></p>
<p>The book came at this result by taking the SF and BM as positive (as is shown in the figure).</p>
<p>If I take the direction of SF and BM opposite to what the books takes (i.e. shear force tends to produce an antickw rotation of the element) I get</p>
<p><span class="math-container">$$\frac{dV}{dx}=q$$</span></p>
<p>The relations are different depending on what sign of SF and BM I take. However I have seen the book using the first relation (with a -ve q) even in places where the element is acted upon by -ve shear force (i.e. when the shear force tends to rotate the element antickw)</p>
<hr />
<p>Why the equation <span class="math-container">$$\frac{dV}{dx}=-q$$</span> applies even when the shear force tends to produce an antickw rotation, even though it was derived for clockwise rotation shear force.</p>
| |mechanical-engineering|structural-engineering|stresses| | <p>The reason the convention picks the shear positive when on a differential element it tries to rotate clockwise is that this rotation will produce a positive conventional moment, bending the beam into a smile.</p>
<p>This convention makes the calculation easy and because conventionally a horizontal beam is analyzed from left to right and the positive shear direction is vertically up, this convention works.</p>
<blockquote>
<p>Why the equation
<span class="math-container">$\frac{dV}{dx}=−q$</span>
applies even when the shear force tends to produce an antickw rotation, even though it was derived for clockwise rotation shear force.</p>
</blockquote>
<p>Because if you have a section like this it means either q is applied to head up or <span class="math-container">$V2<V1$</span> then we have <span class="math-container">$\ -(-q)=q\ $</span> and this will cause an anticlockwise spin. But still, if we stay with convention it will work.</p>
<p>you can Pick up any direction as positive as long as you stay consistent throughout the length of the beam.</p>
| 49163 | Relationship between shear force and load intensity |
2022-01-08T13:54:46.080 | <p>I am trying to create a prototype for a gravity energy storage system, and I need to release potential energy stored in a heavy load of mass <em><strong>M</strong></em>, raised to a height *<strong>H</strong>.</p>
<p>I intend to use the mass traversing the length, to drive a piston, which pushes upon a hydraulic fluid to turn the P.E into K.E.</p>
<p>I am at a loss as to what mechanism I can use to efficiently (and simply) translate the descent into a vertical movement of the piston - since the piston will be contained in a sealed cylinder.</p>
<p>I have included a rather crude sketch below:</p>
<p><a href="https://i.stack.imgur.com/vmNyT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vmNyT.png" alt="Mechanism" /></a></p>
<p>Where:</p>
<ol>
<li>The parallelogram represents the load bearing surface,</li>
<li>The two dark circles represent the piston/plunger</li>
</ol>
<p>My question is this: Is there a mechanism that allows a plunger/piston to be moved along an axis <strong>WITHIN</strong> a cylinder, by means of external force applied <strong>OUTSIDE</strong> the cylinder containing the piston?</p>
<p>I think what I'm looking for is some kind of telescopic mechanism.</p>
| |mechanical-engineering|mechanisms|hydraulics|pistons| | <p>Energy storage is a fairly common feature in hydraulic systems. Storing energy via nitrogen compression in accumulators is far more cost efficient than storing via mass/gravity. Personally I would just use something off the shelf. Here are a few popular brands:<br />
<a href="https://ph.parker.com/us/en/piston-style-accumulators-parker-a-series" rel="nofollow noreferrer">https://ph.parker.com/us/en/piston-style-accumulators-parker-a-series</a><br />
<a href="https://www.boschrexroth.com/en/xc/products/product-groups/industrial-hydraulics/topics/cylinders/large-hydraulic-cylinders/products-and-features/hydraulic-piston-accumulators/index" rel="nofollow noreferrer">https://www.boschrexroth.com/en/xc/products/product-groups/industrial-hydraulics/topics/cylinders/large-hydraulic-cylinders/products-and-features/hydraulic-piston-accumulators/index</a><br />
<a href="https://www.hawe.com/en-us/products/product-search-by-category/hydraulic-accessories/accumulators/hps/" rel="nofollow noreferrer">https://www.hawe.com/en-us/products/product-search-by-category/hydraulic-accessories/accumulators/hps/</a><br />
<a href="https://www.hydac.com/shop/en/hydraulic-accumulators" rel="nofollow noreferrer">https://www.hydac.com/shop/en/hydraulic-accumulators</a><br />
I'm highlighting piston type accumulators since you specifically asked for that style. Other varieties such as diaphragm type or bladder type are usually cheaper per kJ of storage. Piston type accumulators are really for specialty applications such as extreme pressure (700+ bar), corrosive fluids that dissolve bladder materials, narrow installation dimensions, horizontal installation (bladder type service life is best with vertical installation), etc.</p>
<p>Note that with any type of energy storage, there are many safety considerations regarding how to dissipate energy in machine failure situations. Don't simply throw together an accumulator + pump + hoses. You need additional valves for pressure relief, accumulator dump, pump outlet check valve, etc. Further reading -
<a href="https://www.machinerylubrication.com/Read/30331/hydraulic-accumulators-dangers" rel="nofollow noreferrer">https://www.machinerylubrication.com/Read/30331/hydraulic-accumulators-dangers</a></p>
| 49164 | Is there a mechanism that allows this kind of piston/plunger movement? |
2022-01-09T09:00:38.050 | <p>I asked this in the physics stack but hopefully will get more interest here. So I understand that when driving on a curved path, the inner wheel must travel a shorter distance than the outer wheel. But <strong>why</strong>?</p>
<p>As the car begins to turn, both wheels want to continue rolling at the current speed in linear motion due to conservation of angular momentum. If the wheels are being driven by the engine, then the torque is equal to rolling resistance on both wheels so both maintain a constant speed. If the car is coasting, then both wheels are braked by rolling resistance by equal amounts. In order to change the speed of the inner and outer wheels, there must be additional torque(s), which must be different for the two wheels. What is this torque, and where does it come from? Assume that both wheels are rolling without slipping.</p>
| |mechanical-engineering|automotive-engineering|car| | <p>This video went a long way to help me understand how differentials work and how resistance on one wheel translates to motion on the other.</p>
<p><a href="https://www.youtube.com/watch?v=yYAw79386WI" rel="nofollow noreferrer">Around The Corner - How Differential Steering Works (1937)</a></p>
| 49180 | When a car with an open differential turns, why does the inner wheel slow down and the outer wheel speed up? |
2022-01-10T13:56:43.593 | <p>1 kg of air in a piston-cylinder apparatus can exchange heat only with a reservoir maintained at 300 K.</p>
<p>When 10 kJ of work is done on the air, its state is asserted to change from 1 bar 300 K to 2.5 bar, 310 K.</p>
<p>(a) What is the entropy change of the air?</p>
<p>(b) What is the heat transfer from the air?</p>
<p>for part a,</p>
<p>s = c<sub>p</sub>ln(T<sub>2</sub>/T<sub>1</sub>) - Rln(p<sub>2</sub>/p<sub>1</sub>)</p>
<p>so the entropy change of the air is s = 1.004ln(310/300) - 0.287ln(2.5/1)</p>
<p>change in entropy for air = -0.23 kJ/K</p>
<p>For part b,</p>
<p>I am using the equation ΔS = ΔQ/T(average)</p>
<p>ΔQ = -0.23 x 305 = -70.15 kJ of heat transferred from the air.</p>
<p>My textbook says the correct answer for part b is -2.82 kJ/K</p>
<p>I am struggling to understand how to arrive at this answer and I would appreciate it if someone could explain. Thanks.</p>
| |thermodynamics|homework| | <p>I wouldn't bother trying to solve this problem, because the problem doesn't make sense as written. There are two issues:</p>
<p>I.</p>
<p>If the system is in contact with a heat bath at 300 K, then its final temperature should be 300 K, not 310 K.</p>
<p>I suppose the idea could be that the compression is fast enough that the rate at which thermal energy is generated is greater than the rate at which it can flow out to the bath, and that the specified final state is a temporary non-equlibrium state, and that one is calculating the approximate entropy change between an equilibrium and non-equilibrium state. [I say approximate because the non-equilibrium state doesn't have a well-defined entropy.] But, if so, that should have been mentioned explicitly.</p>
<p>II.</p>
<p>The 10 kJ specified for work can't be right. As you may know, the minimum possible compression work is that done with a reversible process. Using that, let's leave aside the details of 300 K vs. 310 K, and try to get a ballpark constraint on the work.</p>
<p>1 kg of air, at an average molecular weight of 28.96 g/mol, is 33.4 moles.</p>
<p>The reversible work to compress 33.4 moles of air at 300 K from 1 bar to 2.5 bar is:</p>
<p><span class="math-container">$$w = - n R T \ln \frac{V_f}{V_i} = - n R T \ln \frac{p_i}{p_f}=- n R T \ln \frac{1}{2.5} = 76.3 kJ$$</span></p>
<p>[With constant T and n, <span class="math-container">$\frac{V_f}{V_i} = \frac{p_i}{p_f}.$</span>]</p>
<p>This value, which is the work for a reversible process, gives us an approximate lower bound for what the compression work could be (playing with the difference between 300 K and 310 K isn't going to change things much). Thus the 10 kJ value specified in the problem, which is nearly an order of magnitude lower, clearly makes no sense.</p>
<p>Sure, you could get the work down to 10 kJ (indeed, down to 0) by cooling the air to a low temperature, letting it contract, stopping the piston, and heating the air back up. But that's not what the problem specifies.</p>
| 49199 | I am struggling to get the right answer regarding the quantification of entropy |
2022-01-10T18:45:41.023 | <p>According to google searches, the Jinping Underground Laboratories are the "deepest" building or buildings constructed, reaching 7900 feet (2400 metre) below the surface.... However, the surface in question is a mountain. While that does classify as underground, it highlights a flaw in the question of the deepest underground building.</p>
<p>I can't seem to find the lowest building in the world though, or the deepest in relation to depth within the earth's crust.</p>
<p><strong>What is the lowest point below sea level that we have built where a human can go?</strong>
I imagine this is likely another laboratory. But where would a building of this description be?</p>
| |civil-engineering|architecture| | <p>The <a href="https://www.kiddoperations.ca/en/Pages/home.aspx" rel="noreferrer">Kidd Mine</a> in Ontario, Canada: per the Wikipedia <a href="https://en.wikipedia.org/wiki/Kidd_Mine#Depth" rel="noreferrer">article</a>, it is "the deepest accessible non-marine point on Earth" at "2,733 metres (8,967 ft) below sea level". I found this from the Wikipedia article on <a href="https://en.wikipedia.org/wiki/Extremes_on_Earth#Subterranean" rel="noreferrer">Extremes on Earth</a>, which differentiates between depth from the surface and depth below sea level, and also between an actual mine vs. a bore hole.</p>
| 49207 | What is the lowest point below sealevel that we have built where a human can go? |
2022-01-11T19:39:41.907 | <p>Consider a rectangular cross sectional solid beam, fixed at one end and a uniform shear force applied to the other end.</p>
<p>Below shows the side view for this beam.</p>
<p><a href="https://i.stack.imgur.com/vujNV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vujNV.png" alt="enter image description here" /></a></p>
<p>Now, I change the support area as shown below. How should the force distribution be now?</p>
<p><a href="https://i.stack.imgur.com/Yuu3M.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Yuu3M.png" alt="enter image description here" /></a></p>
<p>In the first image, above the neutral axis we already know that the structure will be in tension while below the neutral axis, the structure will be in compression. The net upwards tensile force and net bottom compressive forces are equal, and when combined together, the resultant is a reaction moment (with no net reaction force along the longitudinal axis, and only net reaction force to counter the shear).</p>
<p>Now, for the second image, again the reaction moment will exist. Since no external force is applied along the longitudinal axis, there won't be any reaction force along that axis. The only reaction force that will exist is the force to counter the shear. So how would the tension and compression forces be distributed (on the cross section marked with blue line) so that they will return a total reaction moment only?</p>
| |finite-element-method|solid-mechanics| | <p>If your question is :</p>
<blockquote>
<p>how should be the force distribution at the cross section right before the fixed support? This was my actual question, actually.</p>
</blockquote>
<p>Since the structure needs to transfer internally a bending moment right before the support there will be a tension layer and a compression layer. The combination of those layers in each cross-section is that creates <em>internally</em> the bending moment that resists the load.</p>
<p>The exact size of the tension layer and the compression layer (they need to be somewhat symmetric, even for the simple case (isotropic material, constant cross-section) is not something that is usually covered in textbooks. IMHO the best way to address that is either experimentally or FEA.</p>
| 49226 | How will the reaction moment on the support be distributed in terms of forces in this model? |
2022-01-14T19:32:32.490 | <p>I have a locking mechanism in mind, that is made with two racks.
Rack 1 is fixed and then we push down rack 2 to lock something. After that a horizontal force is applied to rack 2. I calculated the resulting vertical force as shown in the sketch.
I have two questions:
Is the model accurate enough or do we over estimate the vertical force too much or with other words how much does the friction affect the vertical force?
I summed all the forces on one tooth of rack contact. Does taking into consideration more teeth change things a lot?</p>
<p><a href="https://i.stack.imgur.com/mD8rf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mD8rf.jpg" alt="Sketch of mechanism" /></a></p>
<p><strong>UPDATE</strong></p>
<p>I managed to analyze the forces on the slope. Everything is on the sketch. I also calculated that for the coefficient of friction 0,36, slope is self locking at an angle 20°. Steel on steel is friction 0,12 and we get self locking at around 7°. With self locking there is also no vertical force.</p>
<p>On thing is that I can't answer to myself. Why is my first equation with tangens inncorect? Triangle of horizontal, normal and vertical force there is different than with horizontal, normal and shear force on this sketch. Maybe someone can help me with this.</p>
<p><a href="https://i.stack.imgur.com/w6lU7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w6lU7.png" alt="Sketch" /></a></p>
| |friction| | <p>Assumptions:</p>
<ul>
<li>neglecting forces from gravity (weight of top part)</li>
<li>neglecting possible effects from offset load</li>
</ul>
<hr />
<p>(1) Decompose the external force into orthogonal components:</p>
<p><span class="math-container">$\vec{F_H} = \vec{F_N} + \vec{F_T}$</span></p>
<p>(2) The geometry gives us the magnitudes of the decomposed components</p>
<p><span class="math-container">$F_N = F_Hcos(\alpha)$</span></p>
<p><span class="math-container">$F_T = F_Hsin(\alpha)$</span></p>
<p>(3) Write the stipulated equilibrium condition: that the parts "stick".</p>
<p><span class="math-container">$$F_T = {\mu}F_N$$</span></p>
<p>(4) Solve the above for limiting value of <span class="math-container">$\mu$</span></p>
<p><span class="math-container">$$F_Hsin(\alpha) = {\mu}F_Hcos(\alpha)$$</span></p>
<p><span class="math-container">$$\mu = \frac{sin(\alpha)}{cos(\alpha)} = tan(\alpha)$$</span></p>
<p>(5) The "vertical force" on the bottom part is just the vertical component of <span class="math-container">$F_T$</span></p>
<p><span class="math-container">$$F_Tcos(\alpha) = F_Hsin(\alpha)cos(\alpha)$$</span></p>
| 49271 | Rack on rack vertical force |
2022-01-15T00:39:23.663 | <p>I have been out of SolidWorks CAD modeling for some time. Recently started getting into some more intense CAD modeling in SolidWorks 2021 and noticed that the mouse controls have changed. I tried some internet searching, but I evidently don't know the right keywords.</p>
<p>The old system would allow you to view the model from any angle in 3D space. You could click the center mouse wheel and rotate the mouse clockwise or counter clockwise to index the model. It was a bit non-intuitive to starting out, but powerful for getting the exact view you wanted. This is the best video I found and it was in 2019. Notice how as he rotates the mouse clockwise the model also rotates. <a href="https://youtu.be/FTQZBDIv6gs?t=52" rel="nofollow noreferrer">https://youtu.be/FTQZBDIv6gs?t=52</a></p>
<p>Currently in 2021, the model always returns to the upright position and this ability to rotate the model to any angle has been removed (at least in the default settings). I put together this youtube video real quick of the navigation issue I am having:
<a href="https://youtu.be/wmfQ4pdB-j8" rel="nofollow noreferrer">https://youtu.be/wmfQ4pdB-j8</a></p>
<p>Is there some setting I need to turn off/on? Is there some new and improved way to view models from any angle in space that supersedes the old method?</p>
| |solidworks| | <p>I've found the setting which you inadvertently have selected - it's called "Rotate About Scene Floor" Simply right click off the model in the graphics area, and uncheck this option.</p>
<p><a href="https://i.stack.imgur.com/9ybH9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9ybH9.png" alt="Rotate About Scene Floor" /></a></p>
<p>The reason one might choose this - is that it makes the camera rotation behave in a more similar way to some rendering packages. Users who are used to that environment will be as frustrated trying the 'normal way' as you were this way.</p>
| 49276 | How to switch back to the previous SolidWorks mouse control for rotating a model? |
2022-01-15T22:16:59.850 | <p>I'm looking for examples of frictional materials that have a coefficient of kinetic friction that is as close to equal to their coefficient of static friction as possible.</p>
<p>What is an example of that kind of material (on Earth)? Is there a database of real-world materials that can tell me about this?</p>
| |materials|friction|material-science|surface-modelling| | <p>Steel/Teflon - <span class="math-container">$\mu_s = \mu_k = 0.27$</span></p>
<p>Glass/Teflon - <span class="math-container">$\mu_s = \mu_k = 0.1$</span></p>
<p>Ice/Ice - <span class="math-container">$\mu_s = \mu_k = 0.01$</span></p>
<p>See table <a href="https://mechguru.com/machine-design/typical-coefficient-of-friction-values-for-common-materials/" rel="nofollow noreferrer">here</a>.</p>
| 49289 | Examples of Materials with a Coefficient of Kinetic Friction Very Close to or Equal to Coefficient of Static Friction? |
2022-01-16T23:11:55.583 | <p>(Cross-posting from statistics stackexchange)</p>
<p>Say we have a permanent-magnet DC motor that roughly obeys the system equation
<span class="math-container">$$\ddot{x}(t) = \alpha \dot{x}(t) + \beta u(t) + \gamma $$</span>
where <span class="math-container">$x(t)$</span> is the displacement of the rotor and <span class="math-container">$u(t)$</span> the applied voltage at time <span class="math-container">$t$</span>.</p>
<p>Say we wish to determine the values of <span class="math-container">$\alpha, \beta$</span> and <span class="math-container">$\gamma$</span> experimentally. If we can only directly measure <span class="math-container">$x$</span> and not <span class="math-container">$\dot{x}$</span> or <span class="math-container">$\ddot{x}$</span>, how should we go about estimating these parameters from a set of timeseries measurements of <span class="math-container">$u$</span> and <span class="math-container">$x$</span>?</p>
<p>One naive approach is to compute the derivatives through some central finite difference scheme, and then perform an OLS regression - but it is unobvious how the derivative calculation interacts with the regression. Additionally, I have found in practice that this suffers from a significant amount of regression dilution if the test is allowed to run too long at steady-state (the derivatives vanish here, and so all that's left is the noise).</p>
<p>Is there any more "complete" method for analyze systems like this that handles the differentiation as part of the construction of the regression model? Is there a good theory of correlations between derivatives of time-series data?</p>
| |motors|statistics|linear-systems|system-identification| | <p>Another approach that has not been mentioned here, but which is applicable also to nonlinear systems, is to use a gradient-based optimization scheme over a simulation. Say that you simulate the (potentially nonlinear) system
<span class="math-container">$$
\begin{aligned}
\dot x &= f(x, u, p) \\
\hat y &= h(x, u, p)
\end{aligned}
$$</span>
where <span class="math-container">$p$</span> denotes the parameters of the model. Then form a cost function <span class="math-container">$c(y_m, \hat y)$</span>, where <span class="math-container">$y_m$</span> denotes measured outputs and <span class="math-container">$\hat y$</span> are estimated / simulated outputs, for example
<span class="math-container">$$c(y_m, \hat y) = \sum_{t} (y_m(t) - \hat{y}(t))^2$$</span>
Estimating the parameters then amounts to minimizing <span class="math-container">$c$</span> with respect to the parameters <span class="math-container">$p$</span>. This can be done by, e.g., following the negative gradient
<span class="math-container">$$-\dfrac{dc}{dp}$$</span>
where <span class="math-container">$c$</span> depends implicitly on <span class="math-container">$p$</span> due to <span class="math-container">$\hat y$</span> depending on <span class="math-container">$p$</span>.
This procedure is called "gradient descent". Obtaining this gradient by hand can be tedious and error prone, especially since it involves simulating the ODE <span class="math-container">$\dot x = f(x, u, p)$</span>, but many modern programming languages have good support for automatic differentiation which makes obtaining the gradient trivial even for very complicated simulators such as PDE and DAE solvers. For a system with a very small number of parameters like this, simply running a gradient free optimization algorithm would likely do the trick.</p>
<p>This approach replaces the need for differentiating the data by instead using an (ODE) integrator. This approach often has good low-frequency properties, indeed, if you have an error in your model and you integrate, the error tend to be inflated.</p>
<p><a href="https://sensitivity.sciml.ai/dev/ode_fitting/optimization_ode/" rel="nofollow noreferrer">This documentation</a> has a large number of tutorials where automatic differentiation (AD) is used to optimize cost functions that involve simulating different kinds of differential equations. For linear system, AD can be used to find the parameters of a linear state-space model using the prediction error method (PEM), documentation available <a href="https://baggepinnen.github.io/ControlSystemIdentification.jl/dev/ss/#PEM-(Prediction-error-method)" rel="nofollow noreferrer">here</a> (full disclosure, I'm the author of this open-source software). PEM has the added benefit of estimating not only the model parameters, but also a noise model / Kalman filter at the same time. Apart from the open-source software linked to above, PEM is available also in widely used commercial system-identification packages.</p>
| 49298 | System identification of a simple motor with only position measurements |
2022-01-17T06:24:16.550 | <p>Good morning,</p>
<p>I'm trying to solve this problem of kinematics of an eight bar linkage through the analytical method.
I don't want that you perform the calculation, only I would like to have a advice how to divide the mechanism in analyzable subchains...
In the image below I'm able to recognise:</p>
<ol>
<li>A0-A-C-B is an engine</li>
<li>E0-E-F is another engine</li>
</ol>
<p>Actually I have calculated the position of point C on link 3, but then I have been stuck...</p>
<p>My problem is to get the position and so the velocity and acceleration of point D. E0-E with C-D seems a shaper chain but choosing so the calculations complicates a lot.
<a href="https://i.stack.imgur.com/tG9K7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tG9K7.png" alt="enter image description here" /></a></p>
<p>Thanks</p>
<p>Regards</p>
| |applied-mechanics|kinematics| | <p>Finally I have found the time to write down the solution and the Visual Basic Macro in excel to implement the Newton Method
Once you have found easily Xc and Yc...</p>
<p><a href="https://i.stack.imgur.com/57MBw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/57MBw.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/ekq7t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ekq7t.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/6A8yS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6A8yS.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/UqmO8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UqmO8.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/YCS1h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YCS1h.png" alt="enter image description here" /></a></p>
| 49303 | Eight bar Linkage (Analytical solution) |
2022-01-17T18:54:38.763 | <p>I am working on some code that will handle the arm of a robot, and I want to be able to hold the arm in place against the force of gravity. The arm is in 2 parts just like a human arm and I intend to hold the arm in place by setting a certain power to the motor that controls the main arm (equivalent to a human shoulder). Both arms can rotate in any direction around their respective attachment points, but I have code implemented that will keep the second part of the arm from falling down due to gravity. <a href="https://i.stack.imgur.com/CkJX7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CkJX7.png" alt="Arm attached to the robot" /></a> My problem is that I do not know how to calculate the torque of the shoulder part of the arm when the second part is added. So how would I calculate the torque of the main arm at angle 1 with the added force of the second part of the arm at angle 2?</p>
| |mechanical-engineering|mathematics|robotics| | <p>"I want to be able to hold the arm in place against the force of <strong>gravity</strong>... Both arms <strong>can rotate in any direction</strong> around their respective attachment points."</p>
<p>Your system is unstable, which is caused by the fact that there is no fixity against rotation. Note that the "probable force" in the lower sketch can be induced by the deflection shown in the first sketch, and can occur in both X & Z directions. Once the motion starts, the vertical arm 2 will act as a pendulum in the free swing. The resultant (demand on) torques are not predictable without modifying the connection joints or taking into consideration of the kinetic motion and dynamic effects.</p>
<p><a href="https://i.stack.imgur.com/BRX5f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BRX5f.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/meCvo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/meCvo.png" alt="enter image description here" /></a></p>
| 49315 | Calculating Torque With a Second Arm Attached to the Main Arm |
2022-01-19T10:19:38.530 | <p>I am searching for a <strong>Formal</strong> <strong>Definition</strong> of bandwidth given the transfer function of a system, or the bode plot.</p>
<p>Is it the frequency <span class="math-container">$\omega_b$</span> in which
<span class="math-container">$$20log|H(j\omega_b)| = -3\text{dB}$$</span>
or
<span class="math-container">$$20log|H(j\omega_b)| = 20log|H(j0)| - 3\text{dB}$$</span></p>
<p>And if there is a definition without dB, I would like to know about it.
Thanks in advance!</p>
<p><em>EDIT-homework</em>: Application of definition: In which interval is <span class="math-container">$\omega_b$</span> contained? <a href="https://i.stack.imgur.com/hqRyf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hqRyf.png" alt="enter image description here" /></a></p>
| |electrical-engineering|control-engineering|control-theory|homework| | <p><strong>Observation:</strong> Bandwidth is defined as the frequency range <span class="math-container">$[\omega_1 \ \omega_2]$</span> over which the control of the system is "effective". Usually, <span class="math-container">$\omega_1 = 0$</span> and then, by definition, <span class="math-container">$\omega_2 = \omega_B$</span> is the <em><strong>bandwidth</strong></em>.</p>
<p><strong>Definition:</strong> The (closed-loop) <strong>bandwidth</strong>, <span class="math-container">$\omega_b$</span>, is the frequency where the norm of the sensitivity function, <span class="math-container">$|S(j\omega)|$</span>, first crosses the <span class="math-container">$-3\text{dB}$</span> line from below. This means that:
<span class="math-container">$$ \big|S(j\omega_B)\big| = \frac{1}{\sqrt{2}} = 0.707 \approx -3\text{dB} $$</span></p>
<p>The <em><strong>bandwidth</strong></em> in terms of the closed loop transfer function <span class="math-container">$T(s)$</span>, <span class="math-container">$\omega_{BT}$</span>, is the highest frequency at which the norm of the closed loop transfer function, <span class="math-container">$|T(j\omega)|$</span>, crosses the <span class="math-container">$-3\text{dB}$</span> line from above. However, this is usually a poor indicator of performance and is mathematically stated as:
<span class="math-container">$$ \big|T(j\omega_{BT})\big| = \frac{1}{\sqrt{2}} = 0.707 \approx -3\text{dB} $$</span></p>
<p>Τhe <em><strong>gain crossover frequency</strong></em>, <span class="math-container">$\omega_c$</span>, is the frequency where the norm of the loop transfer function, <span class="math-container">$|L(j\omega)|$</span>, first crosses the <span class="math-container">$0\text{dB}$</span> line from above. And mathematically:
<span class="math-container">$$ \big|L(j\omega_c)\big| = 1 = 0\text{dB} $$</span>
These three frequencies, and for systems which have a phase margin of the order <span class="math-container">$PM < 90^o$</span>, are related with one another by the following inequality:
<span class="math-container">$$ \omega_B < \omega_c < \omega_{BT} $$</span>
All of the above definitions and frequencies of the system can be used as specifications for controller design and especially when using the loop shaping design method, where the designer tries to impose certain behaviour to the loop transfer function <span class="math-container">$L(s)$</span>. As a last comment, I would like to write down the relations between the functions mentioned: <span class="math-container">$L(s), S(s), T(s)$</span>. Suppose we have a plant described by <span class="math-container">$G(s)$</span> and which is controlled using the controller <span class="math-container">$K(s)$</span>. Then:</p>
<p><span class="math-container">$$ L(s) = G(s)\cdot K(s) \rightarrow \text{Loop Transfer Function} $$</span>
<span class="math-container">$$ S(s) = \frac{1}{1+L(s)} \rightarrow \text{Sensitivity Function} $$</span>
<span class="math-container">$$ T(s) = \frac{L(s)}{1+L(s)} \rightarrow \text{Closed Loop Transfer Function} $$</span>
<span class="math-container">$$ P(s) = 1 + L(s) \rightarrow \text{System's Characteristic Polynomial} $$</span>
Note that for the loop transfer function the multiplication should be <span class="math-container">$G(s)\cdot K(s)$</span> because when dealing with multivariable systems (MIMO) this is how you get the proper loop transfer matrix. Only if both matrices <span class="math-container">$G(s), K(s)$</span> are <strong>diagonal</strong> then the multiplications <span class="math-container">$G(s)\cdot K(s) = K(s)\cdot G(s)$</span> produce the same results. However, for single-input single-output systems (SISO), it doesn't matter.</p>
| 49328 | Definition of Bandwidth as angular frequency |
2022-01-20T12:26:54.423 | <p>Problem -</p>
<p><a href="https://i.stack.imgur.com/q7lXd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q7lXd.png" alt="enter image description here" /></a></p>
<hr />
<p>I have interpreted "by hinging them together at B" as following -</p>
<p><a href="https://i.stack.imgur.com/g0ZMs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/g0ZMs.jpg" alt="enter image description here" /></a></p>
<p>Now if this is the case, then applying a load on the right beam will have no effect, in my understanding, on left beam. Right beam will press against the left support B and in return the support will apply a reaction force on the right beam.</p>
<p>The left beam is simply resting on the pin so there shouldn't be any "stressing" in the left beam and hence the BM at A should be zero.</p>
<p><a href="https://i.stack.imgur.com/2ixGI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2ixGI.jpg" alt="enter image description here" /></a></p>
<p>As in the diagram above, the beam BC applies reactions P/4 and P/4 (a total of P/2) on the pin. However these reactions will be balanced by surface reactions, so the beam AC as such should experience no force.</p>
<p>But turns out the answer is <span class="math-container">$PL/2$</span></p>
<p>What's going wrong?</p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam|homework| | <p>An internal hinge is a special device used to link two beam segments. Similar to the typical pin support, it can generate reactions in the direction opposite to the load (applied force), with the condition that structural equilibrium must be maintained within the support, <span class="math-container">$\sum F_x = 0$</span> and <span class="math-container">$\sum F_y = 0$</span>.</p>
<p><a href="https://i.stack.imgur.com/ttveO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ttveO.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/vDE8c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vDE8c.png" alt="enter image description here" /></a></p>
<p>Both the systems above are considered as "structurally determinate", however, it becomes "structurally indeterminate to the first degree if an inclined force is applied. I leave it to you to figure out why.</p>
<p><a href="https://i.stack.imgur.com/UPj8s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UPj8s.png" alt="enter image description here" /></a></p>
| 49343 | A problem on beams |
2022-01-21T20:29:01.660 | <p>I am trying to design a VTOL system that uses a servo to rotate a propeller and the attached motor so that it can provide thrust when it is at 0 degree and lift when it is at 90 degree. However, I understand that when the propeller is rotating, it behaves like a gyroscope and if the propeller starts to rotate, there is got to be stresses developed at both the propeller blades and the propeller shaft connections. But I have no idea how to start forming an equation.</p>
| |moments|propulsion|servo| | <p>In order to calculate the mass moment of inertia accurately your best bet is using the functions of a CAD system (solidworks, inventor, onshape etc).</p>
<p>If all you want is an an approximation and you can afford the assumption that each blade of the propeller is a long bar with made from a single material (or that the density is uniform along the length of the blade), then you could calculate the mass moment of inertia about the rotation axis as a sum of all3 blades and the hub.</p>
<h2>blade</h2>
<p>With the above assumption (i.e. long bar of uniform density) then the mass moment of rotation about the rotation axis is:</p>
<p><span class="math-container">$$I_{b} = \frac{1}{3} m_b \cdot r_b^2$$</span></p>
<p>where</p>
<ul>
<li><span class="math-container">$r_b$</span> is the length of the blade</li>
<li><span class="math-container">$m_b$</span> is the mass of the blade</li>
</ul>
<p><a href="https://i.stack.imgur.com/qOPOR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qOPOR.png" alt="enter image description here" /></a></p>
<p>**Figure: blade approximation as a rotating bar (source: <a href="https://en.wikipedia.org/wiki/List_of_moments_of_inertia" rel="nofollow noreferrer">wikipedia</a>) **</p>
<h2>hub</h2>
<p>the blade can be approximated as a rotating disk of mass <span class="math-container">$m_h$</span>, and radius <span class="math-container">$r_h$</span>, and in that case the mass moment of inertia would be:</p>
<p><span class="math-container">$$I_h =\frac{1}{2} m_h \cdot r_h^2$$</span></p>
<p><a href="https://i.stack.imgur.com/IEOWf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IEOWf.png" alt="enter image description here" /></a></p>
<h2>Total</h2>
<p>The total mass could be approximated by:</p>
<p><span class="math-container">$$I_h + n_b\cdot I_{b} = \frac{1}{2} m_h \cdot r_h^2 + n_b\cdot \frac{1}{3} m_b \cdot r_b^2 $$</span></p>
<p>where: <span class="math-container">$n_b$</span> is the number of blades on the propeller.</p>
| 49367 | How to calculate the moment of inertia of a propeller |
2022-01-22T11:15:25.817 | <p>I have an issue with chemical plating of Iron, the component is immersed in Methyl Ethyl Ketone for a time span and then Zinc galvanized plating starts to corrode. I am not understanding what measures should I take. I am not at all into chemical engineering I am embedded engineer and trying to sell my products to companies.<a href="https://i.stack.imgur.com/MibUn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MibUn.jpg" alt="enter image description here" /></a></p>
<p>My advisor said I should get them alkaline coated.</p>
<p>This effect takes place after 3 days and there are no immediate results to the test.</p>
<p>Thank you
You guys are awesome.</p>
| |chemical-engineering| | <p>Ritmesh, just to be clear: are the iron parts cleaned in MEK <em>before</em> zinc coating, or are the zinc-coated iron parts corroding <em>after</em> being exposed to MEK during service as drum clamps? I ask because the the usual process for pre-zinc dip coating is an initial acid wash and rinse/dry, not an MEK cleaning dip.</p>
<p>Also please note 35μm of zinc is not very thick and may not be able to cover up all the pits & scratches in the iron.</p>
<p>Also I notice that the corrosion is <em>pitting corrosion</em> associated with the sheared edges of these parts which suggests pinholes in the zinc in the roughened zones. Two things to try: a <em>tumble deburring</em> of the parts before zinc coating to remove any sharp edges, and a thicker zinc coat.</p>
<p>Finally, if the problem is pinholing then you should be able to rapidly produce it by dipping the parts into a warm, well-oxygenated weak acid solution with a little salt in it and then watching closely for gas evolution at the zinc surface, where the pinholes are. You can also do the pinhole test by assembling an electrochemical cell where one electrode is copper and the other is your zinc-coated part and the solution is a weak acid. this will resolve pinholes in seconds if you do it right.</p>
| 49376 | Reaction of Components with Methyl Ethyl Ketone makes it corrode. How to prevent? |
2022-01-22T18:02:16.147 | <p>I'm making pocket holes to connect wood together with a jig, the Kreg 720. The 720 works by moving at an angle (which I'm trying to measure) to ensure the hole is drilled higher on thicker pieces.</p>
<p>The jig makes a hole like this at a 15 degree angle.
<a href="https://i.stack.imgur.com/bUICZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bUICZ.png" alt="enter image description here" /></a></p>
<p>The 720 works like this:</p>
<p><a href="https://i.stack.imgur.com/JQByp.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JQByp.gif" alt="enter image description here" /></a></p>
<p>I'm trying to calculate the length of the drill bit which I can set with a stop. The 720 has stops included for common wood, but I like to be exact and I need to understand the geometry of how the 720 works. I zeroed out an angle meter and found that the case is 57 degrees, but I took some measurements at the top and bottom and through taking the tangent, figured out that the sliding block moves closer to 58 degrees.</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th></th>
<th>s</th>
<th>d</th>
</tr>
</thead>
<tbody>
<tr>
<td>high point (mm)</td>
<td>43.9</td>
<td>54</td>
</tr>
<tr>
<td>low</td>
<td>10.1</td>
<td>0</td>
</tr>
<tr>
<td>accuracy</td>
<td>0.1</td>
<td>0.1</td>
</tr>
<tr>
<td><span class="math-container">$\Delta$</span></td>
<td>33.8</td>
<td>54</td>
</tr>
<tr>
<td><span class="math-container">$\theta$</span></td>
<td>57.95648</td>
<td></td>
</tr>
</tbody>
</table>
</div>
<p>But! I wanted to make sure that the drilling jig block moved at 57 degrees, so I used my caliper to measure the board thickness block at the bottom and the top. That gave me these measurements.</p>
<p><a href="https://i.stack.imgur.com/gjLSI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gjLSI.png" alt="enter image description here" /></a></p>
<p>Per this geometry:
<a href="https://i.stack.imgur.com/wsdtF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wsdtF.png" alt="enter image description here" /></a></p>
<p>From this I'm trying to find <span class="math-container">$D$</span> the length of the shaft of the pilot screw given <span class="math-container">$t_{\text{min}}$</span>, the minimum thickness, <span class="math-container">$t_b$</span> the thickness of the board for pocket holes, <span class="math-container">$t_T$</span> the thickness of the board you are screwing into, <span class="math-container">$\theta$</span>, the angle of the jig, and <span class="math-container">$s$</span>, the length of the shaft of the screw.</p>
| |measurements|geometry|wood|calibration| | <p>This is what your measurements tell me:</p>
<p><a href="https://i.stack.imgur.com/9XeAt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9XeAt.png" alt="enter image description here" /></a></p>
<p>This is the suggested measurement you shall take if you are interested in the incline of the cutting plane. (Note, "a" is a fixed point on the moving drill bit)</p>
<p><a href="https://i.stack.imgur.com/DOoIJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DOoIJ.png" alt="enter image description here" /></a></p>
| 49380 | Figuring out Geometry for Kreg 720 |
2022-01-24T11:26:29.563 | <p>For long freight trains and those that will be climbing to stations at higher altitudes, an extra or two locomotives are attached to the front. I've always wondered why.</p>
<p>For argument's sake, if there are 30 bogeys each weighing 10 tons, the three engines are pulling a combined 300 tons. Each must be applying the exact same pull else if one is pulling harder than it is effectively taking on all the work with the other two idling. Plus, the load on the first coupling is 300 tons, on the second 290, and so on.</p>
<p>On the other hand, if locomotives are placed after every 10 bogeys, then each is pulling 100 tons only.</p>
| |mechanical-engineering|rail| | <p>To optimise efficiency with multiple locomotives, none of the driving wheels must slip. This means that coordination is necessary between the locomotives.</p>
<p>Passenger trains have been worked as <em>multiple units</em> for more than a century with driving motors (electric or diesel with mechanical transmission) distributed along the length of the trains, as you suggest. Coordination is by a <em>train line</em> which is a multi-wire connection communicating the throttle opening (and gear, where appropriate) between the units.</p>
<p>This is relatively easy for passenger trains because they are short (compared with freight trains) and all the cars are normally owned by the same company, so it's easy to ensure compatability between the units.</p>
<p>Freight trains are much longer, the wagons may be owned by different companies and are often of widely varying types. They are also frequently added and removed during a journey. All this makes it much simpler to put all the locomotives together at the front (as was done in steam days when each locomotive had its own crew and coordination was achieved by whistle signals) rather than try to run a train line to remote locomotives.</p>
<p>Now that there is only one crew, it's still convenient to have all the locomotives together for fault-finding and start-up / shutdown, which is normally not available remotely.</p>
<p>More recently (I think in the last 20-30 years), certain countries permit distributed power with a radio link between the head end and a locomotive in the middle of the train. It's important that the remote locomotive behaves appropriately <em>when</em> the radio link is lost (this often happens in long curves in rock cuttings).</p>
<p>In steam days, one or more <em>banking</em> locomotives would push trains up steep grades from the rear without coupling, then drop back at the summit. This is now considered dangerous because of the difficulty of coordinating the head-end and banking locomotives (for example, if it's necessary to make an emergency brake application, the banking locomotive will probably keep pushing).</p>
| 49405 | Why are multiple locomotives attached only to the front for larger trains? |
2022-01-25T06:52:40.720 | <p>I am looking for sources on the carbon intensity of wind power.</p>
<p>Concrete doesn't directly emit CO2 but the process to make it does in fact emit it. Because concrete is made of aggregate, which contains carbon, doing any chemistry on it will release that. The sources on the internet imply a ton of concrete releases a similar amount of CO2.</p>
<p>So it is obvious that the manufacture of wind turbines involves emission of CO2. Furthermore, wind turbines might not generate enough power to manufacture concrete for another turbine. Has anyone done the math on how much concrete a turbine needs, how much CO2 this concrete emits, and compared it to the counter factual CO2 savings of the turbine?</p>
<p>Just as some example math, it looks like a 1mw turbine saves "2100 lb per hour". Say it lasts 100k hours (?). That's 100kt co2.</p>
<p>This appears to be at 100% efficiency so a better number might be 1-10kt. Then the thousand tons of concrete produce a similar amount of co2 so it's about 1:1.</p>
| |wind-power| | <p>According to the following <a href="https://www.sciencedirect.com/science/article/pii/S0960148111002254" rel="nofollow noreferrer">study</a>, the values for entire cycle of a 1.8MW-gearless and 2MW-gearbox are the following:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: left;"></th>
<th style="text-align: center;">Units</th>
<th style="text-align: center;">Turbine 2.0 MW-geared</th>
<th style="text-align: center;">Turbine 1.8 MW-gearless</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: left;">1 Total CO2eq</td>
<td style="text-align: center;">t</td>
<td style="text-align: center;">1164</td>
<td style="text-align: center;">578</td>
</tr>
<tr>
<td style="text-align: left;">2 Total cumulative energy requirements</td>
<td style="text-align: center;">GWh</td>
<td style="text-align: center;">3.91</td>
<td style="text-align: center;">2.11</td>
</tr>
<tr>
<td style="text-align: left;">3 Annual energy generated</td>
<td style="text-align: center;">GWh</td>
<td style="text-align: center;">5.98</td>
<td style="text-align: center;">3.27</td>
</tr>
<tr>
<td style="text-align: left;">4 Energy payback time (2)/(3)</td>
<td style="text-align: center;">yr</td>
<td style="text-align: center;">0.65</td>
<td style="text-align: center;">0.64</td>
</tr>
<tr>
<td style="text-align: left;">5 CO2e</td>
<td style="text-align: center;">g/kWh</td>
<td style="text-align: center;">9.73</td>
<td style="text-align: center;">8.82</td>
</tr>
</tbody>
</table>
</div>
<p>The total CO2 equivalent for the lifetime (from manufacturing to dismantling) is in the order of 1000 t of CO2 (measured indirectly as energy required to manufacture the wind turbine). In about 8 months (0.65 of a year) that energy is produced from a working wind turbine (Keeping in mind that the wind turbine works for 20-30 years)</p>
| 49421 | Carbon and wind energy |
2022-01-25T11:56:10.110 | <p>Excuse my terminology if it's wrong. I'd like to know if a venturi causes strain on a pump.</p>
<p>For example, in my drawing. I have an air pump behind the house with a 50mm pipe fitting. On the other side of the house is a pool with air jets with the same 50mm fitting. Best would be to just buy a lot of 50mm pipes and join the two around the house. Problem is that the 50mm is quite expensive. That made me think, would you just drop the side of the pipe in between and open it up again at the end and end up with the same result?</p>
<p>So to questions on this:</p>
<ul>
<li>Would this cause strain on air pump because of it needing to "press the air into a smaller whole"?</li>
<li>Would your air output at the pool jets still feel the same?</li>
<li>Bonus question, would water work in the same way as air in this example?</li>
</ul>
<p>Thanks for any help! This project made me very venturi-curious haha.</p>
<p><a href="https://i.stack.imgur.com/d9YuZ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d9YuZ.jpg" alt="enter image description here" /></a></p>
| |fluid-mechanics|airflow|pipelines|flow-control| | <h3>A smaller pipe restricts the flow</h3>
<p>Think of it like using a coffee stirrer straw to drink versus a big thick straw. Or instead of using a garden hose you used a thin piece of tubing. Or tried to move traffic using a 6-lane highway versus a single track dirt road.</p>
<p>The technical details behind it is that any fluid passing through a tube undergoes friction losses. A large diameter pipe has more volume versus the friction from the walls of the pipe (volume increases with the square of the diameter, circumference linearly with the diameter). Thus a thin pipe has more wall circumference versus volume. This is typically expressed as "head loss," and what you see in the pipe is that pressure goes down the further you go in a pipe. A thin pipe has more head loss.</p>
<p>Now, does it cause "Strain" on a pump? I doubt an air pump cares very much about the flow rate it passes. You can get into problems with a water pump if flow drops too much because the water provides lubrication and cooling. So really the only issue here is the reduced flow from the smaller pipe.</p>
| 49428 | Does a venturi cause strain on a pump |
2022-01-25T15:55:55.273 | <p>In my high school chemistry class I am studying the affect of surface area over the rate of reaction. In this case, using <em>Collision Theory</em> to describe how a high surface area to volume ratio contributes to a greater chance of particles colliding, which leads to more successful collisions per second; given reactants are at room temperature and other variables constant.</p>
<p>Since flour grains (in fine powdered form) is highly flammable through combustion with oxygen and may even explode when particles that are suspended in the air are ignited, could it offer a more energy efficient method of converting from chemical to kinetic energy through a combustion engine? I was thinking that due to flour's high air to fuel character and that it's</p>
<blockquote>
<p>35 times more combustible than coal dust.</p>
</blockquote>
<p>Would it be possible for the expanding combustion gas to push the piston, which in turn rotates the crankshaft and drives the engine?</p>
<p>I am aware that the mechanical components behind this <a href="https://patents.google.com/patent/CN202811062U/en" rel="nofollow noreferrer">'flour driven engine'</a> may not be similar to the traditional combustion engine. However, would the mechanics behind this design be realistic or probable in the future, if so, how might it look like? For example in the form of an air cylinder? What is the air to fuel ratio of four in comparison to volatilised fuel?</p>
| |mechanical-engineering|chemical-engineering| | <p>I'm going to ignore molecular collisions, etc. and stick to the empirical world, which is where we engineers excel.</p>
<p>Energy content of flour is 353 Kcal/100g. This equals 1.477 MJ/100g</p>
<p>Energy content of gasoline is 46.5 MJ/kg, or 4.65 MJ/100g</p>
<p>So, what this tells us is that gasoline has a much higher energy density than flour. In practical terms, this means that a "flour engine" would need to be much physically larger to get the same work out if it. A larger engine is unlikely to be more efficient.</p>
<p>This ignores the practical implications of feeding an engine with an aerosol flour charge. People have tried this with coal for quite a while and have been generally unsuccessful. Coal-water slurry combustion can work, but you have the problem of having to boil off the water, which doesn't help efficiency, and with pollution controls, which are all harder with coal.</p>
<p>Now, <em>efficiency</em> means getting the most work out of a device relative to the energy put into it. Given the 100 years of regular fossil fuel engine development, we'll be at quite a disadvantage in getting our Flour Engine up to an equivalent efficiency. Given all the infrastructure in place to move & store fossil fuels, moving flour around to do the same thing seems like a waste anyway.</p>
<p>The last issue is that moving food production into energy production is terrible for society, resulting in higher food costs. People need food more than transportation. Attempts to make this happen (like with ethanol) are driven by the farm lobby, not by good policy.</p>
<p>So my answer is, maybe you could get it to work, but <strong>why would you want to?</strong></p>
| 49435 | Can carbohydrates in powdered form provide a more 'energy efficient' alternative solution to traditional combustion engines? |
2022-01-26T00:34:03.117 | <p><em>What</em> and <em>where</em> are they, and what do they <em>look like</em>?</p>
<p>Do all transportations with roundabouts use them in some form?</p>
<p>This is different from <a href="https://engineering.stackexchange.com/q/11611/36705">What's the purpose of a 'burger' lane in a roundabout?</a>, which is very specific to <em>why</em>. This is an open floor to describe where and what they are, since not all transportation systems will have these.</p>
| |civil-engineering|traffic-intersections| | <p>if your question is how they look like then:</p>
<p><a href="https://i.stack.imgur.com/1TPkg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1TPkg.png" alt="enter image description here" /></a></p>
<p><strong>figure: source <a href="https://www.hulldailymail.co.uk/news/hull-east-yorkshire-news/hulls-new-a63-hamburger-roundabout-3044314" rel="nofollow noreferrer">hulldailymail</a></strong></p>
<p>or</p>
<p><a href="https://i.stack.imgur.com/ywAUO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ywAUO.png" alt="enter image description here" /></a></p>
<p><strong>figure: source <a href="https://wiki.openstreetmap.org/wiki/Throughabout" rel="nofollow noreferrer">openstreet wiki</a></strong></p>
<p>They are also known as <a href="https://en.wikipedia.org/wiki/Roundabout#Through_roundabout" rel="nofollow noreferrer">"through roundabouts"</a></p>
<p>As to their name I suppose that the names comes from the fact that it looks like a hamburger. I.e. the two buns are the green islands and the two lanes of the road are the cheese and bacon slices.</p>
| 49452 | "Burger" lanes: What are they, where are they found, and what do they look like? |
2022-01-26T08:31:05.833 | <p>There are various dashboards (for example, <a href="https://www.energydashboard.co.uk/live" rel="nofollow noreferrer">here</a>) that show the current energy generation mix and the overall CO2 emissions (gCO2/kWh).</p>
<p>In the UK renewables can make up a large proportion of the total, but it is never 100% - there is always some fossil fuel generation (from natural gas and occasionally coal).</p>
<p>Naively, I feel I should charge my electric car when the CO2 emissions are low (typically on days when wind generation makes up a greater proportion of the total energy mix).</p>
<p>And a recent article in the New Scientist (<a href="https://cacm.acm.org/news/256584-smart-scheduling-for-big-computing-tasks-cuts-emissions-up-to-a-third/fulltext" rel="nofollow noreferrer">Smart Scheduling for Big Computing Tasks Cuts Emissions Up to a Third</a>) also concluded that CO2 emissions are lower if you plug in when renewable sources account for a greater proportion of the energy mix.</p>
<p>On the other hand, plugging my car into the grid presumably increases the load, which will be supplied by non-renewable generation (since renewables don't yet cover 100% of demand).</p>
<p>So, it seems to me that maybe it's worth avoiding times when coal-fired power stations contribute to the mix, but otherwise it perhaps doesn't matter whether gas supplies 10% or 90% of the total: the power needed to charge my car will always come from burning more gas.</p>
<p>The question: can I reduce my CO2 emissions by charging the car when renewable energy generation makes up a larger proportion of the mix? (Knowing that renewables never contribute 100% and the shortfall is covered by burning fossil fuels).</p>
| |electrical-engineering|energy-efficiency|renewable-energy| | <p><strong>tl;dr: Very windy days might be worse but the data isn't great. Charging in the late afternoon is your best bet to reduce <em>marginal</em> emissions.</strong></p>
<hr />
<p>From the question:</p>
<blockquote>
<p>On the other hand, plugging my car into the grid presumably increases the load, which will be supplied by non-renewable generation (since renewables don't yet cover 100% of demand).</p>
</blockquote>
<p>What you have described is the <a href="https://sustainability.stackexchange.com/q/7112/3379">marginal energy</a> -- basically, which power source will be adjusted in response to a change in demand. This is determined by a complicated mix of physics, engineering, market forces, and, of course, weather, however by analyzing historical data it's possible to determine how this varies with time, without needing to understand all the complicated factors that influence it.</p>
<p>The paper <a href="https://doi.org/10.1016/j.enpol.2016.11.012" rel="nofollow noreferrer">"Marginal greenhouse gas emissions displacement of wind power in Great Britain"</a> provides an analysis of the marginal emissions factors considering both time of day and month of the year.</p>
<p>The charts below show the marginal emissions factors (MEF) for 2009 (in red) through 2014 (in black). You can see that as wind power capacity is added to the grid over time, the overall MEF has dropped, but there are still daily and seasonal variations.</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/ahop5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ahop5.png" alt="Diurnal and seasonal marginal emissions factors in the UK, 2009 to 2014" /></a></p>
</blockquote>
<p>Specifically, <strong>it would be best to charge during the late afternoon in December, when the MEF is lowest.</strong></p>
<p>Another interesting result from the paper is that as the wind output to the grid in any moment increases, the MEF also increases:</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/CMEbVm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CMEbVm.png" alt="Marginal emissions factor as a function of wind output in the UK, 2009 to 2014" /></a></p>
</blockquote>
<p>This actually indicates that charging could be best when there's a medium amount of wind -- not on the windiest days. The paper speculates that the reason for this trend is that high wind output correlates with high system demand, meaning that most gas plants are already online, and a higher proportion of marginal energy is actually coming from coal plants.</p>
<p>However, this paper is from 2014, and there have been significant changes to the grid makeup since then. From <a href="https://www.gov.uk/government/statistics/uk-energy-in-brief-2021" rel="nofollow noreferrer">UK Energy Brief, 2021</a>, you can see that coal capacity ("conventional steam") has reduced by nearly half, while renewable (mostly wind) has nearly doubled.</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/0aYVyl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0aYVyl.png" alt="UK electricity capacity, 1996 to 2020" /></a></p>
</blockquote>
<p>Further, the UK government forecasts the <em>long-run</em> marginal emissions factor for the overall grid, out to 2100. "Long-run" just refers to a permanent change, more useful for long term planning and analysis. The data is available here: <a href="https://www.gov.uk/government/publications/valuation-of-energy-use-and-greenhouse-gas-emissions-for-appraisal" rel="nofollow noreferrer">Green Book supplementary guidance: valuation of energy use and greenhouse gas emissions for appraisal </a>. I put together a chart showing how the emissions factor is expected to change over time:</p>
<p><a href="https://i.stack.imgur.com/tle5m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tle5m.png" alt="UK long-run marginal emissions factor over time" /></a></p>
<p>From this we can see that as renewable penetration on the grid increases, the timing of when power is used becomes less and less important; by 2040 or so, it won't really matter at all.</p>
<hr />
<p>Unfortunately I wasn't able to find any updates to this research since 2014, and the National Grid's <a href="https://carbonintensity.org.uk/" rel="nofollow noreferrer">very fancy carbon intensity API</a> doesn't account for marginal emissions at all. But there is some <strong>other related research:</strong></p>
<ul>
<li><a href="https://doi.org/10.1016/j.isci.2021.103499" rel="nofollow noreferrer">"Reducing the life cycle environmental impact of electric vehicles through emissions-responsive charging"</a> looks at how smart chargers could utilize MEF signals to reduce carbon emissions, concluding that overnight charging in the UK has the greatest potential.</li>
<li><a href="https://doi.org/10.1016/j.jclepro.2020.124766" rel="nofollow noreferrer">"Carbon efficient smart charging using forecasts of marginal emission factors"</a> compares use of MEF to average emissions factors (AEF) in Germany for deciding when to charge an EV, and finds that using the AEF would actually increase carbon emissions in some cases.</li>
</ul>
| 49457 | Is it better to charge my electric car on windy days? |
2022-01-26T17:29:46.343 | <p>Consider two slabs - Slab A and Slab B, insulated on "latereal" faces as shown, initially at the same temperature, and having identical dimensions. The slabs at t= 0 are brought in contact with two heat reservoirs (on left and right) at temperatures <span class="math-container">$T_1$</span> and <span class="math-container">$T_2$</span>. Slabs have the same thermal conductivity but different specific heats, with <span class="math-container">$c_A > c_B$</span></p>
<p>Since specific heat of A > that of B I argue that the temperature profiles at any instant of time t, would be as follows:</p>
<p><a href="https://i.stack.imgur.com/Etnxc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Etnxc.jpg" alt="enter image description here" /></a></p>
<p>i.e. since <span class="math-container">$c_A > c_B$</span> A will have a hard time raising it's temperature than B. As a result the temperature gradients in A will be smaller (in magnitude) than in the case of B. This would mean that the heat transferring to A from the left reservoir in any time dt is smaller in A than in B. Furthermore, the rate of heat transfer in intermediate layers will also be lower in A than B. I've often read that a higher specific heat restricts thermal diffusion, could this be one way of explaining it why?</p>
| |mechanical-engineering|heat-transfer|thermal-conduction| | <p>You are right. What you are referring to maybe the property called <strong>Thermal diffusivity, <span class="math-container">$\alpha$</span></strong></p>
<p><span class="math-container">$$\alpha =\frac{k}{\rho C_p}$$</span></p>
<ul>
<li>Cp = specific heat</li>
<li>k = thermal conductivity</li>
<li><span class="math-container">$\rho $</span>= density
So Thermal diffusivity is inversely related to specific heat.</li>
</ul>
<p>However in your example as soon as slab A reaches the T2,
a bit after slab B, on the right side they both transfer heat at the same rate because they have the same thermal conductivity.</p>
| 49466 | Heat Diffusion and Specific heat |
2022-01-26T22:18:19.077 | <p>This problem can be solve in 2 ways either I solve it with vectors which would be relatively painful and more time consuming and the other faster way is algebraically but I faced a problem when trying to find the points of intersection with the axes:</p>
<p>After I calculated <span class="math-container">$R_x=403.3$</span> N , <span class="math-container">$R_y=-131.81 N$</span> and Moment about G(origin): <span class="math-container">$$M_G=-460cos(15)*0.47+100*0.59+120cos(70)*0.47-120sin(70)*0.19+100+135=88.89 N$$</span></p>
<p>Then I said that the sum of moments of forces about G = The moment of the resultant force about G</p>
<p><span class="math-container">$$R_x*y+R_y*x=88.98$$</span>
<span class="math-container">$$\therefore -131.81x+403.3y=88.98$$</span></p>
<p>Now when I plug <span class="math-container">$ y=0 :x=0.675m=675mm$</span></p>
<p>And when <span class="math-container">$x=0$</span> : <span class="math-container">$y=-0.2207m=-220.7mm$</span></p>
<p>Apparently there is no answer with the signs that I got ,What did I do wrong here.
<a href="https://i.stack.imgur.com/evEpF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/evEpF.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|applied-mechanics|statics|solid-mechanics|moments| | <p>The idea was correct, and all your calculations were correct. You only neglected to consider the sign of the generating moment, i.e. in the 2D case the moments of a force are given by the following equation (notice the minus) :</p>
<p><span class="math-container">$$M= -F_x\cdot y + F_y\cdot x$$</span></p>
<p>So for horizontal forces</p>
<ul>
<li>when a positive horizontal force is applied on a positive y distance then the resulting moment is negative</li>
</ul>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;"></th>
<th style="text-align: center;">Positive y</th>
<th style="text-align: center;">Negative y</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">Positive <span class="math-container">$F_x$</span></td>
<td style="text-align: center;"><a href="https://i.stack.imgur.com/vodlL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vodlL.png" alt="enter image description here" /></a> <br> <strong>- M</strong></td>
<td style="text-align: center;"><a href="https://i.stack.imgur.com/OSoix.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OSoix.png" alt="enter image description here" /></a> <br> <strong>+M</strong></td>
</tr>
<tr>
<td style="text-align: center;">Negative <span class="math-container">$F_x$</span></td>
<td style="text-align: center;"><a href="https://i.stack.imgur.com/SEr3Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SEr3Z.png" alt="enter image description here" /></a><br><strong>+M</strong></td>
<td style="text-align: center;"><a href="https://i.stack.imgur.com/21qUz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/21qUz.png" alt="enter image description here" /></a> <br> <strong>-M</strong></td>
</tr>
</tbody>
</table>
</div>
<p>Similarly, <strong>vertical forces</strong>:</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th style="text-align: center;"></th>
<th style="text-align: center;">Positive x</th>
<th style="text-align: center;">Negative x</th>
</tr>
</thead>
<tbody>
<tr>
<td style="text-align: center;">Positive <span class="math-container">$F_y$</span></td>
<td style="text-align: center;"><a href="https://i.stack.imgur.com/v691j.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v691j.png" alt="enter image description here" /></a><br> <strong>+ M</strong></td>
<td style="text-align: center;"><a href="https://i.stack.imgur.com/yGH6S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yGH6S.png" alt="enter image description here" /></a><br> <strong>-M</strong></td>
</tr>
<tr>
<td style="text-align: center;">Negative <span class="math-container">$F_y$</span></td>
<td style="text-align: center;"><a href="https://i.stack.imgur.com/NsXOR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NsXOR.png" alt="enter image description here" /></a><br> -M</td>
<td style="text-align: center;"><a href="https://i.stack.imgur.com/ZONXw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZONXw.png" alt="enter image description here" /></a><br> + M</td>
</tr>
</tbody>
</table>
</div>
<p>So when you were calculating the x coordinate, only the y component of the force generates moment (i.e. $R_y), so what you should have calculated was:</p>
<p><span class="math-container">$$M_G = +R_y\cdot x \Rightarrow$$</span>
<span class="math-container">$$x = +\frac{M_G}{ R_y}=+\frac{88.98}{ −131.81}= -220.65 mm$$</span></p>
<p>and similarly for the y calculation</p>
<p><span class="math-container">$$M_G = -R_x\cdot y \Rightarrow$$</span>
<span class="math-container">$$y = +\frac{M_G}{ R_x}= - \frac{88.98}{ 403}= -675 mm$$</span></p>
| 49470 | Is there a way to solve this complex statics problem algebraically? |
2022-01-27T01:14:00.723 | <p>My teacher and I argued about whether a seismic event causing a force in itself or being an inertial response of the system. As I read in Chopra and Paz, the effect of an earthquake can be interpreted either as a force or as an inerntial response, either way yielding in the same results. I think it can be seen as both and my teacher says its an inertial response and nothing else. So I told my teacher and the class that we both were correct, but my teacher argued (and threatened) that he's right and I'm wrong. I wanted to see other peoples opinion.</p>
| |structural-engineering|civil-engineering|earthquake-engineering| | <p>Don't confuse reality with equivalent mathematical representations.</p>
<p>A classic way of thinking about dynamic systems is through <a href="https://en.wikipedia.org/wiki/D%27Alembert%27s_principle" rel="nofollow noreferrer">D'Alembert's principle</a>, which allows us to convert a dynamic system into a static system under inertial forces.</p>
<p>I'm no expert in dynamic systems and have never heard of Chopra and Paz (the advantages of living in a country in the middle of a tectonic plate: no need to worry about earthquakes!). But I wouldn't be surprised if they're basically just applying D'Alembert's principle, converting the earthquake into equivalent forces.</p>
<p>So, are you correct that "the effect of an earthquake can be interpreted either as a force or as an [inertial] response, either way yielding in the same results"? Yes, you are. But that doesn't mean that the earthquake itself can be thought of as a force or an inertial response with equal validity.</p>
<p>The earthquake is the earth literally moving under your building's feet. Your structure's inertia then causes it to behave a certain way, generating certain forces throughout.</p>
<p>You can absolutely <em><strong>model</strong></em> that inertial response as a force, but that doesn't take away from the fact that --- <em><strong>in reality</strong></em> --- you're dealing with an inertial response.</p>
<p>You're free to question "if they give the same result, then who cares?" The answer is "pedants", but that doesn't take away from the fact that they aren't wrong.</p>
| 49473 | In a structure, an earthquake causes a force in itself or it causes an intertial response in the system? |
2022-01-27T12:18:26.993 | <p>At the risk of every woodworker in the world snickering at me, I'm curious how strong my wood joints are and want to know how much weight a table supports. I'm an aero engineer who ended up in EE, but I should remember how to do this, but some basic searches didn't show me how to answer:</p>
<ul>
<li>Given a thread depth, <span class="math-container">$d$</span> in wood with a some strength properties (I think modulus of elasticity (MOE) and modulus of rupture (MOR) would be relevant), what is the strength of the joint at varying screw lengths?</li>
<li>How would I quantitively compare a butt-joined screw from a pocket screw for different loads?</li>
</ul>
<p>I know one way to do this would be to use FEM. On the other hand there are probably some empirical tables with rough parameters, but I'm looking for a middle ground of some basic rules/math that help me get a feel for the strength of stuff I'm building. I am eager to get smart on how to model this with Fusion360, where I do have explicit modeling of the screws in my cad models.</p>
<p>Btw, <a href="https://link.springer.com/article/10.1007/s00107-021-01668-4" rel="nofollow noreferrer">this paper</a> may have some insight and there are sites <a href="https://awc.org/codes-standards/calculators-software/connectioncalc" rel="nofollow noreferrer">like this</a> with calculators. It would be great to find some resources that break down the design trades.</p>
| |mechanical-engineering|strength|fasteners| | <p>American Wood Council has standards including design examples for all sorts of wood screws, bolts, lag-screws, etc.</p>
<p>It addresses many factors such as species of the wood or woods if different species are to be connected, their density, moisture, and others. <a href="https://awc.org/pdf/education/des/AWC-DES345-ConnectionDesignExamples-171019.pdf" rel="nofollow noreferrer">source</a></p>
<p>Here is an example of designing the allowable withdrawal of a lag screw.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/Q9VpL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q9VpL.png" alt="lag screw" /></a></p>
| 49480 | Screw Calculations/Math |
2022-01-27T12:32:22.620 | <p>Let there be a spring as shown. And this is stretched to point C and released, then force exerted by spring on mass in region X2 is "- kx" . But This springs gains kinetic energy at mean position and moves till P. Could you please tell me the force expression (force produced by spring on mass) in X1 region i.e., at pt. A at some distance x. Assume suitable variables and massless mass attached to spring. <a href="https://i.stack.imgur.com/PRqDc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PRqDc.jpg" alt="Attached image for reference" /></a></p>
| |mechanical-engineering|springs|kinematics|acceleration|force-measurement| | <p>I think you meant to say the spring is massless, correct me if I'm wrong.</p>
<p>Force on the mass by string if we assume down direction as positive ( to follow your sign -kx pulling up) is <span class="math-container">$kx, \ $</span> it is pushing the mass down.</p>
<p>At the same time, the force exerted by the mass on the spring is <span class="math-container">$-kx$</span>.</p>
<p>The mg component is permanently pulling the mass down and as far as inertia it doesn't come into play. All it does is moves the equilibrium position down by an amount <span class="math-container">$$x_{equilibrium}= \frac{mg}{k}$$</span></p>
<p>And the mass is experiencing two forces <span class="math-container">$mg+kx$</span></p>
<p>Usually, we set the reference point at the end of the string where it attaches to the mass by the hook, not at the middle, but it doesn't make a difference.</p>
| 49481 | Inertial Force expression in spring |
2022-01-27T13:51:28.290 | <p>In the figure below it is stated that the friction force should always be greater than the tangential force in order to prevent slipping between two frictional wheels. My question is that if we apply Newton's second law to that contact point won't the tangential force resultant point upwards with the direction of the friction since the tangential force is larger and thus the wheels have an opposite rotating direction?<a href="https://i.stack.imgur.com/HF5n0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HF5n0.jpg" alt="1" /></a></p>
| |mechanical-engineering|gears|dynamics|kinematics|machine-design| | <p>The concept behind</p>
<blockquote>
<p>the friction force should always be greater than the tangential force</p>
</blockquote>
<p>is that the <em><strong>maximum</strong></em> friction force should be greater to the force that causes the rotation of the wheel. If you try to apply more torque, then there will be slipping.</p>
<p>This is very similar to the following concept:</p>
<p><a href="https://i.stack.imgur.com/H0GAv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H0GAv.png" alt="enter image description here" /></a></p>
<p>If a small force <span class="math-container">$F_A$</span> is applied, then the friction force will only be equal to that force <span class="math-container">$F_A$</span> and the box will not move. If <span class="math-container">$F_A$</span> exceeds the <em><strong>maximum</strong></em> friction force then the box will <strong>slide</strong>.</p>
<hr />
<p><strong>Important Sidenote</strong>:
That sliding motion is what we usually consider as motion, and sometimes get confused in the case of the gear the OP presented. I.e. the rotation again is considered motion. <em>However</em>, in the case of the gears there is <em>no sliding</em> -- i.e. for the contact point of gear A and gear B the relative velocity is zero, or in another way the move in unison.</p>
<hr />
| 49482 | Frictional wheels without slipping (gear system) |
2022-01-31T19:12:48.140 | <p>The book I'm following for Strength of materials states that, in a beam with symmetric cross section (about the plane of bending) and in pure bending, the moment of cross sectional area about the neutral axis is zero, i.e.</p>
<p><span class="math-container">$$\int_A y\,dA=0$$</span></p>
<p>where <span class="math-container">$y$</span> is the perpendicular distance of an area <span class="math-container">$dA$</span> from the neutral axis.</p>
<p>The book then says that, this could only happen when the neutral axis passes through the centroid of section, and I don't understand why.</p>
<p>I have a feeling this might be very trivial to ask, but please bear with me.</p>
<p><img src="https://i.stack.imgur.com/xFCzb.jpg" alt="enter image description here" /></p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam| | <p>So we know that the moment of the cross sectional area about the neutral axis is zero that is,</p>
<p><span class="math-container">$$\int_A y\,dA=0$$</span></p>
<p>where y is the perpendicular distance from the neutral axis of an area <span class="math-container">$dA$</span></p>
<p>Let the origin of the coordinate system lie at the center of Neutral Axis. The <span class="math-container">$y$</span> in the above equation is the distance measured along the y axis of this coordinate system.</p>
<p><a href="https://i.stack.imgur.com/6FtN4.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6FtN4.jpg" alt="enter image description here" /></a></p>
<p>The centroid C, of this cross sectional area will lie somewhere on the y axis. The y coordinate of the centroid is given by,</p>
<p><span class="math-container">$$\bar y= \frac{\int_A y\,dA}{\int_A dA}$$</span></p>
<p>since <span class="math-container">$$\int_A y\,dA=0$$</span></p>
<p><span class="math-container">$$\bar y=0$$</span></p>
<p>This suggests that the centroid is at <span class="math-container">$y=0$</span>, i.e. on the Neutral axis. Conversely, the neutral axis passes through the centroid of the cross sectional area.</p>
<p><a href="https://i.stack.imgur.com/isTQx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/isTQx.jpg" alt="enter image description here" /></a></p>
| 49543 | How does one conclude from this, that neutral axis passes through the centroid? |
2022-01-31T21:07:03.073 | <p>If I am to make a block using the materials for making paper, what would its compressive strength be? I am interested in not the stacked paper test, but using the same material in a block form since I do not believe they will fail the same way (or if you think otherwise, please explain your logic behind your deductions).</p>
<p>In terms of the material of paper, I am pretty flexible. The ideal will be a range of papers with shorter fibers and less interfiber connections, and papers with longer fibers and more interfiber connections.</p>
| |strength| | <p>The answer depends on the fiber length, diameter, chemical makeup, moisture content, direction of fiber alignment relative to the applied stresses, and what <em>binder</em> is added to make the fibers stay stuck together.</p>
<p>Without knowledge of all these things, the compressive strength of a fiber block cannot be predicted.</p>
| 49546 | If I am to make a block using the materials for making paper, what would its compressive strength be? |
2022-01-31T23:49:50.067 | <p>What is the correct way to draw a free body diagram (FBD) for the following cylinder? Assume the cylinder and disc have been twisted to an angular deflection of <span class="math-container">$\theta$</span>. Further assume that we are drawing the FBD immediately after releasing the disc, so static equilibrium <em>does not</em> hold.</p>
<p><a href="https://i.stack.imgur.com/b1QBF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b1QBF.png" alt="enter image description here" /></a></p>
<p>I know there is a restorative force that grows as <span class="math-container">$\theta$</span> gets larger. But how should I write it? Do I need to talk about the shear stress? Or can I simply write a restorative force <span class="math-container">$F_t$</span> tangent to the surface of the disc at the bottom?</p>
<p>For example, would this be a correct FBD?</p>
<p><a href="https://i.stack.imgur.com/fk2f9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fk2f9.png" alt="enter image description here" /></a></p>
<p>Would it be correct to also write the moment or torque that the force at the surface creates? Or is this redundant and would imply, according to the diagram, double the moment acting on the disc?</p>
<p><a href="https://i.stack.imgur.com/2Tq00.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Tq00.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|dynamics|springs| | <p>First of all the problem described --i.e. twisting then releasing -- seem like a dynamics problem (more than a statics one).</p>
<p>Additionally, (and this will become more relevant below) is that the problem as it is described is a <em>torsional spring</em>.</p>
<p>There are a few subtle points.</p>
<ol>
<li>it is not described how the beam ends up in the <span class="math-container">$\theta$</span> angular displacement. Is it a force like the one below?</li>
</ol>
<p><img src="https://i.stack.imgur.com/fk2f9.png" alt="https://i.stack.imgur.com/fk2f9.png" /></p>
<p>if that is the case then you'd be expected to have some lateral displacement, so I suspect it is moment that is applied.</p>
<p>This means that upon release the FBD of the beam has only a reaction force from the support.</p>
<ol start="2">
<li>Upon release the entire beam is still "feeling" reaction force, and is starting to untwist. This is very much the same way that when you extend a spring and then you let it go, the reaction force is proportional to the position. (Actually the problem as it is described is a torsional spring.)</li>
</ol>
<p>So the force will be proportional to the</p>
<ol start="3">
<li><p>Usually, in the beam as twisting and the disk (due to its massively greater second moment of area) as a rigid component. If you can further assume that the beam is massless then the problem approximates very well that of the spring and mass.</p>
</li>
<li><p>Finally, in this type of dynamics problems it is best to draw two types of diagrams. The FBD which sums up the forces, and the <em>kinetic diagram</em> with all the inertial forces. E.g. for a sliding block the FBD and kinetic diagram are shown below.</p>
</li>
</ol>
<p><a href="https://i.stack.imgur.com/zKPHt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zKPHt.png" alt="enter image description here" /></a></p>
<p>Essentially they reflect the two sides of the equation <span class="math-container">$\sum F = m\cdot a$</span>.</p>
| 49554 | How to draw a free body diagram of a cylinder under torsion? |
2022-02-01T13:04:39.773 | <p>Language = C++</p>
<p>There are 2 types of statements I noticed for taking in output.</p>
<p>1)</p>
<pre><code>int x=3,y=4;
if(x>y)
cout<<“x is greater than y”<<endl;
else
cout<<“y is greater than x”<<endl;
</code></pre>
<ol start="2">
<li></li>
</ol>
<pre><code>int a,b,c;
cout<<“Enter 3 numbers”;
cin>>a>>b>>c;
</code></pre>
<p>In the 2nd statement , it is not written <<endl; but only “;” .
I noticed there is no error in compiling. What is the difference between writing the two ?</p>
| |computer-engineering| | <p><code>endl</code> is the end of the line keyword/constant for the C++ compiler.</p>
<p>The only difference, is that when you are prompted to enter numbers you will continue from the same line.</p>
<p>if <code>cout<<“Enter 3 numbers”<<endl;</code> is used then the numbers will be entered below.</p>
<p><a href="https://stackoverflow.com/questions/213907/stdendl-vs-n">diffrence between endl and \n.</a></p>
| 49559 | What is the difference b/w <<endl; & ;? |
2022-02-01T13:50:50.543 | <p>I would like to measure air flow speeds in a duct system for dust collection in a workshop. Relevant air speeds in the ducts are around 10-30 m/s. The ducts are circular tubes with a diameter of 160 mm and thus the Re numbers will be around 100,000 to 300,000.</p>
<p>A pitot-static tube (a.k.a prandtl tube) in combination with an accurate-enough manometer appears to be the conventional measurement equipment for this application. My question is, do I really need a profesional-quality pitot-static tube for about 80-120 USD, or would it be sufficient with a 5 USD tube seemingly intended for RC planes?</p>
<p>In other words, what is the difference between a cheap and an expensive pitot tube? Can we expect substantial differences in performance (e.g., robustness, accuracy, ...), and if so what would be the relevant parameters? Or is it rather about which guarantees are made by the manufacturer? The factor 20 price difference suggests something fundamental could be different.</p>
<p>Does tip geometry play an important role? Is it important that static and dynamic pressure are measured in the same device? Is it important how the holes for dynamic and static pressure are located? Are the professional-quality obviously <em>better</em> in any such parameters?</p>
<p>Could I even do the measurement by measuring the differential pressure between (1) a hose with an opening perpendicular to the flow and (2) a thin, bent brass pipe with its opening oriented against the flow?</p>
<p>I am asking partly because I don't want to waste 100 USD, and partly because I want to understand the engineering considerations beyond the fundamentals of Bernoulli's equation.</p>
| |fluid-mechanics|pressure|measurements|airflow| | <p>The things I know about are:</p>
<ul>
<li>calibration (accuracy, time constants, accuracy variations due to AOA, accuracy variations due to speed/Reynolds number)</li>
<li>proper geometry to minmize variations due to AOA</li>
<li>proper geometry and positioning for static ports to measure undisturbed flow</li>
<li>Things like orifice size affects response time due to Reynold's number/viscosity and what speeds/Reynolds numbers it will be accurate at</li>
</ul>
<p>Remember some of these tubes are lab grade instruments.</p>
<p>You don't need to measure static and dynamic pressure in the same device, but it sure is convenient and theoretically more accurate if you can do it (because otherwise static ports must be placed at another appropriate location where static pressure could also be different.</p>
<p>I think I trust a home made pitot tube more than a pitot-static tube where aerodynamics matter more, You'll probably be fine with a dust collection setup where flow is always parallel and you can be off by 30% and probably won't know the difference. Hobbies tubes just follow TLAR. Which means hobbiest pitot tubes just slap a coned or domed nose on a tube so it looks prettier. Hobbiest Prandtl tubes, in particular, are quite ugly because the construction difficulties of concentric tubes.</p>
<blockquote>
<p><em>Could I even do the measurement by measuring the differential pressure between (1) a hose with an opening perpendicular to the flow and (2) a thin, bent brass pipe with its opening oriented against the flow?</em>"</p>
</blockquote>
<p>I assume you are talking about 90 degree bent probes that enter the flow stream from the duct walls.</p>
<p>Just having a straight open ended tube for static pressure protruding into the duct trips up the air around the static inlet. You need to also bend it pointing into the stream so it is parallel to the flow so as not to disturb it. Then you plug the end with a geometry to not trip the air before to flows down the outside of the parallel tube segment. I think a hemisphere is more tolerant to AOA differences but the sharp end of a tear drop works better if the flow is always head on. I don't know if you can rely on this though because flows have disturbances and hitting the sharp point off angle will trip the flow.</p>
<p>Then you have holes along the length of the parallel segment facing 90 degrees to the flow.</p>
<p>Some Prandtl-tubes have a boundary trip behind the nose after the dynamic inlet for the static port. Others don't. I don't know when you want or do not want to trip it but it's a careful thing so if you have a separate static probe just place it away from disturbances and don't trip it. I think the trip might be there to introduce turbulence to prevent something like a laminar bubble or flow separation on top the static port if it is expected. That should be worse than just having turbulent but attached flow at the static port.</p>
<p>Those are my armchair conclusions after doing the same research a few years ago that you are doing now.</p>
| 49565 | What is the difference between a 100 USD professional pitot tube and a 5 USD hobbyist pitot tube? |
2022-02-01T17:26:32.313 | <p>i am working on my bachelor thesis and I am supposed to do a parameter study within ansys SpaceClaim. Basically I have a cross sectional geometry (on paper) and need to do the following steps:</p>
<ol>
<li><p>Convert it to some parameterised sketch within Ansys</p>
</li>
<li><p>For a given set of parameters, adjust the sketch according to my parameters.</p>
</li>
<li><p>Generate a body from the sketch within Ansys</p>
</li>
<li><p>Compute thermal field</p>
</li>
<li><p>Repeat from step 2</p>
</li>
</ol>
<p>Since I have never worked with Ansys, I do not know how to proceed and which tools I am supposed to use. Could someone here give me a hint which tools would work out best? Is it possible to do parameterised sketches within SpaceClaim and if so, how would I adjust these parameters?</p>
<p>I am very happy for any help!
Greetings Finn</p>
| |ansys| | <p>If you are new to ANSYS and SpaceClaim, then the first thing you should do is just open YouTube, open ANSYS Learning Channel and watch the tutorials for SpaceClaim there. On top of that, watch videos uploaded by other channels as which explains other SpaceClaim features in detail.</p>
<p>I am going to assume you already know how to draw the sketch in SpaceClaim. After making that sketch, you can click the option 'Dimensions' which can be seen in toolbar under 'Sketch' ribbon on top. Then you can click on any edge, circle etc that you have sketched and change its dimension on the bottom left tree under 'General', as shown in the picture below.</p>
<p><a href="https://i.stack.imgur.com/evpmV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/evpmV.png" alt="enter image description here" /></a></p>
<p>The problem with SpaceClaim, as I have observed, is that you cannot parameterize these dimensions. As far as I can recall, you have to manually open the SpaceClaim everytime and then change these parameters manually. There might be an option to parameterize these dimensions but I am not aware of it; maybe you can access online website of SpaceClaim to check how that can be done if it is possible.</p>
<p>For DesignModuler in ANSYS, it is easy to parameterize the dimensions. Just click this option and a <em><strong>P</strong></em> will appear, meaning that this dimension has been parameterized and you can change it directly by opening the parameter table (seen directly on ANSYS Workbench window). There is no need to open the DesignModuler over and over again if the dimension(s) need to be changed.</p>
<p><a href="https://i.stack.imgur.com/FXpCz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FXpCz.png" alt="enter image description here" /></a></p>
| 49571 | Ansys SpaceClaim parameter study |
2022-02-01T19:45:37.507 | <p><a href="https://i.stack.imgur.com/ujEMy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ujEMy.png" alt="enter image description here" /></a></p>
<p>The above beam is loaded by a distributed load per unit length of the referential scale defined by
the vector field <strong><em>q</em> = <em>q(x)</em></strong> and a distributed moment load vector per unit length <strong><em>m</em> = <em>m(x)</em></strong>. As a consequence of the external loads, the beam is deformed into the so-called current state where the external loads are balanced by an internal section force vector <strong><em>F</em> = <em>F(x)</em></strong> and an internal section moment vector <strong><em>M</em> = <em>M(x)</em></strong></p>
<p>The internal force <em><strong>F</strong></em> has normal component <span class="math-container">$N$</span> and shear components <span class="math-container">$Q_y$</span> and <span class="math-container">$Q_z$</span> and the internal moment has the respective components <span class="math-container">$M_x$</span>, <span class="math-container">$M_y$</span>, and <span class="math-container">$M_z$</span>. The equations of equilibrium are as follows:</p>
<p><a href="https://i.stack.imgur.com/3n23b.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3n23b.png" alt="enter image description here" /></a></p>
<p><strong>What I don't understand is why is the cross product of the unit vector <span class="math-container">$i$</span> with <span class="math-container">$dF$</span> equal to zero? Thank you.</strong></p>
<p><a href="http://homes.civil.aau.dk/jc/FemteSemester/Beams3D.pdf" rel="nofollow noreferrer">Source of the tutorial</a></p>
| |beam| | <p>Maybe they are assuming dF is extremely small, for a length change of dx, as compare to dM (which is relatively much larger than dF). So we can put dF to be equal to zero for simplifications.</p>
| 49574 | Moment equilibrium equation of beam |
2022-02-02T11:29:28.083 | <p>Problem:</p>
<p><a href="https://i.stack.imgur.com/mQxdg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mQxdg.png" alt="enter image description here" /></a></p>
<hr />
<p><em>A beam is said to be in pure bending if the bending moment in it remains constant throughout the length</em>.</p>
<p>The problem asks to determine the bending moment in the wire. In the solution of the problem, the textbook uses the formulas which were derived for a beam in pure bending. So I believe the wire is in pure bending, but I don't understand how it is in pure bending.</p>
<p><strong>How the wire is in pure bending?</strong></p>
| |mechanical-engineering|structural-engineering|beam|homework| | <p>A couple of easy ways of producing pure bending are:</p>
<ul>
<li>Applying equal opposing moments at the two ends of the beam, and applying two equal concentrated loads symmetrically spaced.</li>
</ul>
<p>Assuming one of the supports on the digrams is a roller.</p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/rivHe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rivHe.png" alt="pur bending 1" /></a></p>
<p>.</p>
<p><a href="https://i.stack.imgur.com/GqeFg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GqeFg.png" alt="pure bending" /></a></p>
<p>.</p>
| 49580 | How is the wire in this problem in pure bending? |
2022-02-02T20:32:55.580 | <p>I am currently working with the Ansys design modeller and I have a rather complex geometry. In order to compute the required variables for my geometry, I have a very long expression inside my parameter dimension assignment window. Unfortunately it seems to cut off the end of my expressions, resulting in an invalid syntax error for my expression.</p>
<p><a href="https://i.stack.imgur.com/OLyPx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OLyPx.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/zFj0L.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zFj0L.png" alt="enter image description here" /></a></p>
<p>I cannot reduce the amount of characters inside my formula. Is there a way to split my equation into smaller sub equations or to define some intermediate variables which themself depend on my parameters? What is the best solution to my problem?</p>
| |ansys| | <p>Once you declared the parameters in DesignModeler you can access them from the project page. Than you could write a python script for the calculation. Here a short example which accesses all parameters and changes their value according to an expression</p>
<pre><code># get parameter object
parms = Parameters.GetAllParameters()
# print the parameter names and replace their values
for i in range(0,parms.Count):
parmName = parms[i].DisplayText
try:
print parmName
parms[i].Expression = 'Your expression goes here'
except:
continue
# refresh the project
Refresh()
</code></pre>
<p>Of course your expression is limited to IronPythons modules, but on a first glance it looked like your operations are covered in the <code>math</code> module</p>
| 49591 | Ansys Parameter Editor max characters |
2022-02-04T08:06:51.557 | <p>I have this drill and I need one of 6mm, but I don't know how it is called. I think it is used to make a hole round and smooth with high precision. Can anyone identify what the name of this kind of drill is. It is 12mm.</p>
<p><a href="https://i.stack.imgur.com/xzQzt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xzQzt.png" alt="drill" /></a></p>
| |mechanical-engineering|drilling| | <p>What you are <em>describing</em> (for finishing holes to precise dimensions, and possibly therefore what you're actually in need of?) is called a "reamer".
There are many types e.g. for use in a machine or by hand, so I will not provide a link lest it be seen as a recommendation</p>
<p>The item in your photo is a "Cylindrical End-Cutting Rotary Burr" for use with a die grinder. See here for an example: <a href="https://www.zoro.co.uk/shop/cutting-tools/tct-burrs/carbide-rotary-burrs-cylindrical-end-cutting/f/4165?query=Cylindrical+End-Cutting+Rotary+Burr+12mm" rel="nofollow noreferrer">https://www.zoro.co.uk/shop/cutting-tools/tct-burrs/carbide-rotary-burrs-cylindrical-end-cutting/f/4165?query=Cylindrical+End-Cutting+Rotary+Burr+12mm</a></p>
<p>Neither of these things are drill bits.</p>
| 49613 | Identify Drill name or type |
2022-02-04T09:27:23.823 | <p>I'm trying to calculate the type of motor I need to turn something of a certain weight, so I feel the thing I'm trying to calculate is torque, but, when I research how to find torque, I find this "To calculate load torque, multiply the force (F) by the distance away from the rotational axis, which is the radius of the pulley (r). If the mass of the load (blue box) is 20 Newtons, and the radius of the pulley is 5 cm away, then the required torque for the application is 20 N x 0.05 m = 1 Nm. " So that leads me to ask, if I have no pulley, and am directly attaching whatever I want to turn to the axis, then how do I calculate torque without some distance from the axis? If I just multiply it by 0, as in 0m from the axis, then the force needed would be 0, which makes no sense as far as I can see. Is torque even what I should be calculating, and if not, how do I find what I want to know? Thank you!</p>
| |motors|torque|pulleys| | <p>When we talk about an external force trying to rotate a certain object at a pivot point, the moment arm (distance away from the rotational axis) comes into the picture. And the expression for torque would then be equal to,</p>
<p><span class="math-container">$T = F.r$</span></p>
<ul>
<li>T= torque</li>
<li>F = Force acting on the object.</li>
<li>r = moment arm (distance away from the rotational axis)</li>
</ul>
<p>But in your case, you are trying to find the torque of the motor that has to rotate a certain object connected directly to its output shaft. This scenario can be pictured as a force trying to push an object<a href="https://i.stack.imgur.com/2bPee.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2bPee.png" alt="Force acting on an object" /></a>
So the linear acceleration is the actual input, which in turn will give you the value of the force required to that object with that acceleration.
Now coming back to your question. Refer to the image below.
<a href="https://i.stack.imgur.com/rXNwe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rXNwe.png" alt="Torque acting on an object" /></a>
Here, the input would be the angular acceleration which can be calculated using this formula,</p>
<p><span class="math-container">$\theta = \omega_i + \frac{1}{2}\alpha t^2$</span>
if you know the time and angle to be traveled.</p>
<p>With the angular acceleration known to us, we can easily calculate the torque required using the formula,</p>
<p><span class="math-container">$T = I\alpha$</span></p>
<p><span class="math-container">$I$</span> being the mass moment of inertia of the object connected to the motor shaft.</p>
<p>Hope I have answered your question.</p>
| 49615 | Can you calculate torque without any pulleys involved, and if so, how? |
2022-02-04T10:01:21.457 | <p>I have this dynamical system:</p>
<pre><code>A = [ -0.313 56.7 0
-0.0139 -0.426 0
0 56.7 0];
B = [0.232
0.0203
0 ];
C = [0 0 1];
D = 0;
</code></pre>
<p>I'm using the c2d Matlab command to convert it to discrete time. This system is supposed to be stable but why does it behave like this if I use a sample time lower than 1s? With time sample 1s it is stable and converges to 0.</p>
<p>Also, I'm using full state feedback to place its poles at 0.5</p>
<p><a href="https://i.stack.imgur.com/tY97F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tY97F.png" alt="System behaviour " /></a></p>
<p>Thanks in advance</p>
<h2>edit</h2>
<ol>
<li>Matrices given above are for the continuous time system.</li>
<li>The response shown is for the closed loop discrete time system.</li>
<li>The full state feedback is designed for the discrete time system.</li>
<li>There is one pole at origin only, so unitary multiplicity implies stability.</li>
<li>pole placement <code>0.5+0i</code> for the discrete time system.</li>
<li>Full state feedback is designed after doing c2d.</li>
<li>To get full state feedback gain I used <code>-place(Ad, Bd, [0.5 0.501 0.502])</code></li>
<li>I also noticed that placing poles in 0.8 in full state feedback the system converges to 0</li>
</ol>
<p><strong>edit 2</strong></p>
<p>Matlab script:</p>
<pre><code>clear;
close all;
clc;
A = [ -0.313 56.7 0
-0.0139 -0.426 0
0 56.7 0];
B = [0.232
0.0203
0];
C = [0 0 1];
T = 100e-3;
sys = ss(A, B, C, 0);
sysd = c2d(sys, T);
Ad = sysd.A;
Bd = sysd.B;
Cd = sysd.C;
p_des = [0.5 0.501 0.502];
Kr = -place(Ad, Bd, p_des);
N = 100;
x(:, 1) = [0 0 pi/9]';
u(:, 1) = 0;
for i=1:N
if (i<N)
x(:, i + 1) = Ad * x(:,i) + Bd * u(:, i);
end
y(:, i) = Cd * x(:, i);
u(:, i + 1) = Kr * x(:, i);
end
k = 1:N;
plot(k, x');
</code></pre>
| |control-engineering|control-theory|systems-engineering| | <p>What was supposed to be implemented: <span class="math-container">$u_i = Kr\cdot x_i$</span>.
What was actually implemented in the code: <span class="math-container">$u_i = Kr\cdot x_{i-1}$</span>.
This effectively made your third order system into a <em>sixth</em> order system (!) which is unstable. See the figure below.</p>
<p><a href="https://i.stack.imgur.com/q9HN4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q9HN4.png" alt="difference between intention and implementation" /></a></p>
<p>To correct this, the code needs to be slightly modified. Shown below.</p>
<pre><code>%u(:, 1) = 0;
for i=1:N
u(:, i) = Kr * x(:, i);
if (i<N)
x(:, i + 1) = Ad * x(:,i) + Bd * u(:, i);
end
y(:, i) = Cd * x(:, i);
end
</code></pre>
| 49616 | Dynamical system problem |
2022-02-05T08:11:33.850 | <p><a href="https://i.stack.imgur.com/8LRC7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8LRC7.png" alt="enter image description here" /></a></p>
<p>HI all, do these nitro power engine use flywheel with one magnet on one side to time the ignition of the spark plug or do they use other mechanism? Thank you</p>
| |mechanical-engineering| | <p>The ignition source in this type of engine is a <em>glow plug</em>, here is how it works.</p>
<p>The inside of the plug (which screws down into the top of the cylinder head) contains a coil of fine platinum wire. The ends of this coil are connected to a pair of electrical terminals on the outside of the plug. To start the engine, a battery is connected to those terminals, which heats up the wire, and the propeller is spun to rotate the crank and draw a charge of air and fuel into the cylinder. When the piston compresses this mixture, it heats it up and pressurizes it to the point where the hot platinum acts as a catalyst to trigger the explosion of the mixture, and the engine starts to run.</p>
<p>Each explosion maintains the platinum wire hot enough to catalytically trigger the next explosion, and the battery can then be disconnected.</p>
<p>For airplane engines about 30ccs or more in size, a conventional spark plug-and-magneto combination is used, which allows the timing of the spark to be adjusted to allow the engine to run over a greater range of speeds than is possible with glow ignition.</p>
| 49625 | How this engine time its ignition of spark? |
2022-02-06T11:30:43.737 | <p>I am unable to understand how springs and shock absorbers absorb impact. So it would be great if someone could explain how impact is absorbed; and if it could be used to protect someone's leg from beaking when falling from a height.
Thanks in advance.</p>
| |springs|impact| | <p>The classic case of this are the springs and shock absorbers as part of a road vehicle suspension system</p>
<p>The simple answer is that the spring coupled with some sort of lever suspension an object that is subject to the shock is simply a device to allow the shock seen as an impulse like a hammer blow to be absorbed with a modest amount of movement.</p>
<p>The critical part of the mechanism us the damper or shock absorber which is a mechanical device that dissipates energy as the shock is absorbed by the movement. This is often a hydraulic device with a piston that forces a fluid through a narrow nozzle to make the liquid very turbulent this causes the liquid to heat up and this dissipate the energy of the shock.</p>
<p>Without the shock absorber part the spring would just bounce back snd push the infuse energy into a series of oscillations</p>
<p>The clever bit is to know the range of impulsive and to oscillatory shocks that need to be absorbed the range of load on the suspension and the amount of movement that is needed and ro get them matched correctly together to produce the desired results.</p>
| 49634 | Can someone please explain how impact is absorbed? |
2022-02-06T13:29:52.353 | <p>I made this Matlab script to implement a dynamical system with full state feedback and integral action.</p>
<p>Maybe I'm wrong in implementing the integral action because the system diverges.</p>
<p>First I converted the continuous time system into discrete time. Then I extended the state and computed the gain to implement full state feedback and integral action.
Probably there is an error in how the integral action is implemented in <code>for</code> loop.</p>
<p>Shouldn't "integral in discrete time" simply be the previous sample?</p>
<pre><code>clear;
close all;
clc;
A = [ -0.313 56.7 0
-0.0139 -0.426 0
0 56.7 0];
B = [0.232
0.0203
0];
C = [0 0 1];
%conversion to descrete time
T = .1;
sys = ss(A, B, C, 0);
sysd = c2d(sys, T);
Ad = sysd.A;
Bd = sysd.B;
Cd = sysd.C;
%extension
A_ext = [ Ad [0 0 0]'
-Cd 1 ];
B_ext = [Bd
0];
%desidered poles
p_des = [0.5 0.501 0.502 0.503];
K = -place(A_ext, B_ext, p_des);
Kr = K(1:3);
Ki = K(4);
N = 100;
%desidered output
yd = 0.05;
x(:, 1) = [0 0 0]';
u(:, 1) = yd * Ki + Kr * x(:, 1); %(yd - 0) * Ki + Kr * x(:, 1)
x(:, 2) = Ad * x(:, 1) + Bd * u(:, 1);
y(:, 1) = Cd * x(:, 1);
for i=2:N
u(:, i) = (yd - y(:, i - 1)) * Ki + Kr * x(:, i);
if (i < N)
x(:, i + 1) = Ad * x(:,i) + Bd * u(:, i);
end
y(:, i) = Cd * x(:, i);
end
k = 1:N;
plot(k, x');
</code></pre>
<p>Any help would be appreciated.</p>
<p>Thank you</p>
| |control-theory|systems-engineering|systems-design| | <p>The last row of the extended <code>A</code> matrix corresponds to the equation</p>
<p><span class="math-container">$$
x_{\mathrm{integrator}}(i+1) = -C_d x_{\mathrm{rest}}(i) + x_{\mathrm{integrator}}(i) = -y(i) + x_{\mathrm{integrator}}(i)
$$</span></p>
<p>You can use the extended state and matrices inside the <code>for</code> loop as such. Alternatively, if you decide to use the original state matrices inside the <code>for</code> loop as you have done in the sample code, you need to implement the above equation also in the <code>for</code> loop. It will look something like.</p>
<pre><code>x_int_cur = 0
for i = ...
...
% equation corresponding to last row of the
% extended matrix implemented separately
x_int_cur = (yd - Cd * x(:, i)) + x_int_cur
u(:, i) = Kr * x(:, i) + x_int_cur * Ki;
% optional if you want to plot
% the state of the integrator with respect to time later on.
x_int_total_history(:, i) = x_int_cur;
...
end
</code></pre>
<p>Note that, in your sample code, you have used the <em>previous output</em> <code>y(i-1)</code> to feed the integrator. This effectively creates additional states in the system and <em>doesn't</em> match the extended state matrix which uses the <em>current output</em> to feed its integrator.</p>
<p>Also, note that I have not fully checked the modified code above to see if it is exactly matching the extended state matrices. Use it only as a starting point. The main idea is that you have to implement the same extended system equations which you used to <code>place</code> the poles in the <code>for</code> loop also.</p>
| 49636 | Full state feedback with integral action - Matlab script |
2022-02-06T19:22:46.720 | <p>The benefit of true right-handed and left-handed turning lathe tools is obvious: You can get right up against the shoulder or face with them:</p>
<p><a href="https://i.stack.imgur.com/Lnk3X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lnk3X.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/jZElq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jZElq.png" alt="enter image description here" /></a></p>
<p>The benefit of neutral-handed turning lathe tools is also obvious: You can cut in both directions or plunge cut:</p>
<p><a href="https://i.stack.imgur.com/nR3fs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nR3fs.png" alt="enter image description here" /></a></p>
<p>So then what is the advantage of LH and RH turning tools that don't let you get up to the shoulder?
<a href="https://i.stack.imgur.com/22NWy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/22NWy.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/jsPW4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jsPW4.png" alt="enter image description here" /></a></p>
<p>It would seem these would be niche tools for when you need a to be able to plunge cut and get closer to the shoulder with a single tool, but my understanding is that these tools are actually more commonly used than the true LH, RH, and neutral tools. They can't face or get as close to the shoulder as a true LH or RH tool but also can't be as strong as a neutral tool (which can get more material behind the cutting tip).</p>
<p>Are facing and 90 degree shoulders just a lot less important than I think they are?</p>
<p>Images from: <a href="https://www.mcmaster.com/lathe-tools/carbide-tipped-turning-lathe-tools/" rel="nofollow noreferrer">https://www.mcmaster.com/lathe-tools/carbide-tipped-turning-lathe-tools/</a></p>
| |lathe| | <p>So it turns out the tools that can cut a 90 degree shoulder have what is called "negative lead-angle" which results in just the tip cutting and also results in clearance. This allows the tool to face and also turn a 90 degree shoulder.</p>
<p>Conversely, the tools I was asking about in this question have "positive lead-angle". The reason these tools exist is to take advantage of this positive lead angle which cuts along a greater portion of the edge. This results in increased tool life since more of the edge is being utilized and distributes the cutting force along more area which helps reduce certain kinds of failures.</p>
<p>The disadvantage is increased cutting forces for the same depth of cut, increased radial forces, and greater potential for chatter. The increased radial forces result in more part deflection and less cut precision.</p>
<p>What this means is that tools with positive lead-angle tend to best for roughing where tool life and speed matter more than finish and precision.</p>
<p>The reason they are more common than neutral tools is that unless you have extremely large positive lead angles, the neutral tool ends up being too narrow and fragile due to symmetry. The LH and RH tools with positive lead angles don't have this problem since they only cut in one direction so the opposite side can be bulked up with lots of material to support and strengthen the tool.</p>
| 49640 | Usage of Handedness in Lathe Turning Tools |
2022-02-07T11:27:52.213 | <p>Here's an image of the problem:
<img src="https://i.ibb.co/nBtPTSJ/Screen-Shot-2022-02-07-at-5-38-21-AM.png" alt="" /></p>
<p>I'm trying to put the following problem in matrix form (state space) but I don't know how would I put -(g/L)sin(x1) in the A matrix since the x1 value is inside the sin() wave.</p>
<p>Thanks!</p>
| |mechanical-engineering|control-engineering|control-theory|dynamics|kinematics| | <p>Consider the Taylor series expansion</p>
<p><span class="math-container">$$\sin \left(x_1\right) = x_1-\frac{x_1^3}{6}+O\left(x_1^4\right)$$</span></p>
<p>As <span class="math-container">$x_1$</span> gets further away from 0, the correctness of the approximation and that of the linear model with <span class="math-container">$\sin \left(x_1\right) \approx x_1 $</span> decreases.</p>
<p><a href="https://i.stack.imgur.com/Lf46u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Lf46u.png" alt="enter image description here" /></a></p>
| 49646 | How to model the state space matrix of a pendulum? |
2022-02-08T11:20:22.997 | <p>Here is an image of a muffler of an R/C aircraft engine. It's made of aluminium alloys. What do you think about the welding technology which is used to make this muffler? I can't see any welding bead.</p>
<p><a href="https://i.stack.imgur.com/yug8o.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yug8o.jpg" alt="enter image description here" /></a></p>
| |welding| | <p>Cheaper than soldering would be press-fitting the parts together. To do this, at the joint where the (for example) end plug goes into the muffler cylinder, the plug diameter is slightly larger than the inside diameter of the cylinder. When you smash them together with great force, they slide together to form a joint that is nearly impossible to pull apart, yet requires no screws, bolts, welding, soldering or glue to hold it together- and it is fast and cheap to perform in a factory.</p>
| 49660 | What kind of welding technology is used for welding this muffler? |
2022-02-08T11:46:48.433 | <p>I'd like to mount a shelf on the wall using three shelf supports. Where should I place the three supports for best distribution of load across the shelf?</p>
<p>Edit: To clarify, the load will be distributed and I want to optimize for maximum load, not minimum sag.</p>
| |mechanical-engineering|statics| | <p>Assuming a uniformly distributed load throughout the entire shelf, you want to position your supports such that your cantilevers are 40-45% of your central spans.</p>
<p>Now to show my work. For starters, we're obviously going to want our supports to be symmetrically placed. That means we know that one of our supports will be at the middle of the shelf. Shelf supports are usually best described as hinged supports (allowing for a bit of rotation), but since we're dealing with a symmetrical structure, we also know that the rotation at the middle of the shelf must also be zero.</p>
<p>This means we can simplify our structure and think of only one half of it, which we can model as a fixed-and-hinged beam with a cantilever. The image below exemplifies how a three-support beam is identical to two fixed-and-hinged beams:</p>
<p><a href="https://i.stack.imgur.com/0CSQ6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0CSQ6.png" alt="enter image description here" /></a></p>
<p>So now we only have one degree of freedom: where we put that hinged support. And we want to optimize the structure's load-bearing capacity. In this case, there are really two ways for the shelf to collapse: the shelf itself breaks apart (almost always due to excessive bending moment) or the supports break off.</p>
<p>To maximize the shelf's bending moment capacity, we need to minimize its largest bending moment. As it happens, a fixed-and-hinged beam with a cantilever will always have a negative bending moment at the hinged support and either a positive or negative bending moment at the fixed support (our central support in the real beam).</p>
<p>However, it is mathematically impossible for the positive bending moment to be greater (in magnitude) than the negative at the hinged support, so really we just want to position our supports such that the bending moment at both supports are equal.</p>
<p>The math for this is a pain in the neck, so I'm going to skim over it here. The first thing we need to determine is the vertical force generated by the hinged support, which we eventually derive is equal to:</p>
<p><span class="math-container">$$\begin{align}
P &= w\dfrac{a^4 - 4\ell^3a + 3\ell^4}{8(\ell - a)^3} \\
&= w\dfrac{a^2 + 2a\ell + 3 \ell^2}{8(\ell - a)}
\end{align}$$</span></p>
<p>where <span class="math-container">$w$</span> is the applied load; <span class="math-container">$a$</span>, the length of the cantilever; and <span class="math-container">$\ell$</span>, the total beam span (cantilever plus the rest of the beam).</p>
<p>We can then calculate the bending moments at both supports:</p>
<p><span class="math-container">$$\begin{align}
M^-_a &= -\dfrac{wa^2}{2} &&\text{ (hinged support)}\\
M^-_b &= -\dfrac{w\ell^2}{2} - P(\ell - a) &&\text{ (fixed support)}
\end{align}$$</span></p>
<p>Now we just need to define the value of <span class="math-container">$a$</span> such that those two are equal. This is more annoying algebra, but we end up getting that the optimal value is</p>
<p><span class="math-container">$$a = \dfrac{\ell}{1 + \sqrt6} \approx 0.29\ell$$</span></p>
<p>Since <span class="math-container">$\ell$</span> is the entire beam span, we can also conclude that <span class="math-container">$a = \dfrac{a}{1 - a} = \dfrac{1}{\sqrt6} \approx 41\%$</span> of the central span.</p>
<p>We can therefore see that indeed, with these dimensions the maximum bending moment is equal at all critical points (forgiving rounding error), ensuring the shelf is resisting bending moment optimally:</p>
<p><a href="https://i.stack.imgur.com/dDobW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dDobW.png" alt="enter image description here" /></a></p>
<p>However, as I mentioned at the start, we also need to think about how much the supports resist. If our layout is such that the shelf doesn't collapse under bending moment, it may still be overloading one of our supports. So another constraint we have to think about is minimizing the maximum load each support must resist.</p>
<p>As it happens, however, that 41% is still a pretty good choice for this constraint as well. As the image above demonstrates, the load resisted by each of the supports is already quite similar* (~10% difference). I won't get into the math for this case, but <span class="math-container">$a$</span> has to be ~44% of the central span for the supports to all have equal reactions.</p>
<p>Obviously, if your shelf has to deal with other cases such as concentrated loads in only a few positions, etc, then the conclusion may be significantly altered, but an <span class="math-container">$a$</span> between 40-45% of the central span means your shelf will be well prepared to withstand the greatest uniformly distributed load possible.</p>
<hr />
<p>* <sub>The bending moments should be identical and are slightly different merely due to rounding (the cantilever is 41% of the central span, when it should actually be <span class="math-container">$1/\sqrt6$</span>). The support reactions are in fact different.</sub></p>
| 49661 | How should I position three shelf supports for the best distribution of load? |
2022-02-08T11:48:19.747 | <p>I was going through some Vibration examples on how to solve multiple-degree-of-freedom systems and I have noticed that they usually assume a particular solution with just one trigonometric function (sine or cosine depending on the excitation force, and when there is none, they just assume sine) instead of the regular pair</p>
<p><span class="math-container">$$x_p = A \cos{\omega t} + B \sin{\omega t}$$</span></p>
<p>I was wondering <strong>why</strong> they make this assumption and <strong>when</strong> it is allowed (2 questions).</p>
| |vibration| | <p>Although it is not explicitly stated, since you are considered partial solution then the problem considers an external excitation F of some sort.</p>
<p>If I understood correctly the question the reason why <code>a particular solution with just one trigonometric function</code> is assumed is that:</p>
<ol>
<li><p>through the Fourier theorem any time series can be approximated by a summation of many sines/cosines</p>
</li>
<li><p>Using sine or cosine is not a problem because of the identity <span class="math-container">$\cos(\phi) = \sin\left(\frac{\pi}{2} + \phi\right)$</span>, and equally <span class="math-container">$\sin(\phi) = \cos\left( \phi-\frac{\pi}{2} \right)$</span>. So choosing sine or cosine is not a problem</p>
</li>
<li><p>the Laplace transform (which is a very useful tool in those problems) is very simple.</p>
</li>
<li><p>the sum of <span class="math-container">$A\cdot \cos(\omega t) + B\cdot \sin (\omega t) $</span>, can be written as <span class="math-container">$A\cdot \sin(\omega t +\frac{\pi}{2}) + B\cdot \sin (\omega t) \Rightarrow$</span>, and therefore you have the some of two sines at different phases (0 and <span class="math-container">$\frac{\pi}{2}$</span>), which is equal to a sine wave of the same frequency at a different phase.</p>
</li>
</ol>
<hr />
<p>So, the main reason - IMHO - is no.3 because it provides a <em>simple solution</em> (some might giggle at this statement) through the Laplace transform, and that makes it an ideal candidate for textbooks and in-class paradigms.</p>
| 49662 | Particular Solution of Multiple-Degree-of-Freedom Systems |
2022-02-08T15:37:00.640 | <p>I'm designing a battery case for cylindrical li-ion batteries similar to the AA battery cases. We need to avoid spot welding to the cell directly since we want may need to replace some cells later for research purposes.</p>
<p>I'm trying to figure out how much force the spring (battery contact) need to apply to the battery to maintain electrical contact. I'm thinking it could be equal to the weight of the cell, i.e. about 0.5N or 2.8kPa but I couldn't really figure out how other people are doing it.</p>
<p>Would appreciate it if someone could give me some suggestions as to the amount of force it needs. Thank you in advance.</p>
| |mechanical-engineering|design|springs|battery| | <p>Here are some examples from Lee Spring that could be useful. All the AA size springs listed have 1.75 lb nominal load.</p>
<p><a href="https://i.stack.imgur.com/SJvZE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SJvZE.png" alt="foo" /></a></p>
<p>Source: <a href="https://www.leespring.com/battery-springs" rel="nofollow noreferrer">https://www.leespring.com/battery-springs</a></p>
| 49667 | How much force needs to be applied on a battery cell to maintain contact? |
2022-02-09T10:46:50.873 | <p>I am interested in designing a GNC controller for a 3DOF underactuated vehicle that follows a path in 2D space. The two available control inputs are the thrust force in the surge direction and a steering yaw moment for heading change. There are two control objectives here: one, maintain a desired surge velocity <span class="math-container">$u_d$</span> (dynamic task), and two, utilize a guidance law to steer towards a path (geometric task). My question is, if I were to decompose and pursuit these two objectives with independent control laws (standard practice, as is shown later), what would be the overall stability guarantees of the resulting control synthesis?</p>
<hr />
<p><strong>Example:</strong> To illustrate my question further, consider the typical <a href="https://i.stack.imgur.com/K8b5Z.png" rel="nofollow noreferrer">Line-of-sight (LOS) guidance law</a>, <a href="https://www.sciencedirect.com/science/article/pii/S1474667017378096" rel="nofollow noreferrer">introduced by T. Fossen et al.</a> for the control of underactuated vessels.</p>
<p><img src="https://i.stack.imgur.com/K8b5Z.png" alt="Line-of-sight (LOS) guidance law" /></p>
<p>The LOS guidance law says that if an appropriate heading signal <span class="math-container">$ψ_d$</span> is tracked, then the cross-tracking error will be minimized and the path will be followed. Therefore, one can build a GNC controller on this principle, consisting of two control laws, the first being responsible for surge speed control:</p>
<p><span class="math-container">$$\tau u:=−F_u(u,v,r)−k_u(u−u_d)$$</span></p>
<p>where <span class="math-container">$F_u$</span> is a damping force and <span class="math-container">$k_u$</span> a gain factor, while the second control law being responsible for yaw control</p>
<p><span class="math-container">$$\tau_r:=−F_r(u,v,r)+\ddot ψ_d−k_ψ(ψ−ψ_d)−k_r(r−\dot ψ_d)$$</span></p>
<p>where <span class="math-container">$F_r$</span> is a damping moment and <span class="math-container">$k_ψ$</span>, <span class="math-container">$k_r$</span> are gain factors.</p>
<p>Here, control law <span class="math-container">$τ_u$</span> is based on feedback linearization and thus guarantees exponential convergence to <span class="math-container">$u_d$</span>, next, control law <span class="math-container">$τ_r$</span> is globally asymptotically stable, and therefore, it follows mathematically that the overall controller is κ-exponentially stable (see paper).</p>
<hr />
<p><strong>To conclude, and to reiterate the question:</strong> What are the respective requirements for the stability properties of the two control laws, in order to have an asymptotically stable overall system? Can, for example, the aforementioned velocity control law <span class="math-container">$τ_u$</span>, be replaced with an asymptotically stable one?</p>
<p>I'm not quite sure whether my problem falls under the cascade control classification since there is no inner and outer control loop; the reference velocity <span class="math-container">$u_d$</span> is set a priori and the <span class="math-container">$ψ_d$</span> tracking is independently pursued. References or direction ideas that would help me assess stability of such controllers would be appreciated, thanks in advance.</p>
| |control-engineering|control-theory|stability|nonlinear-control| | <p>Okay, so after carefully considering the literature on the topic, the answer is "it depends" - basically what @AJN said: "If the rotational and surge dynamics are uncoupled or weakly coupled, then independent control law synthesis may work. If they are coupled, then the stability is better analysed for the combined system." I should also add that it depends on the types of stability of the respective control laws.</p>
<p>For the case of the control of an underactuated system, the main objective is to show that all dynamics are stabilized given the less-than-necessary control variables. Correspondingly, stability must be approached using a cascaded control systems approach. I refer future readers of this question to the paper <a href="https://www.sciencedirect.com/science/article/pii/S0167691197001199" rel="nofollow noreferrer">"On global uniform asymptotic stability of nonlinear time-varying systems in cascade"</a>.</p>
<p>In the paper, sufficient conditions are given that guarantee that a globally uniformly stable (GUS) nonlinear time-varying (NLTV) system remains GUS when it is perturbed by the output of a globally uniformly asymptotically stable (GUAS) NLTV system, as well as whether two GUAS systems yield a GUAS cascaded system, under some growth restrictions over the Lyapunov function.</p>
| 49679 | Stability properties of controllers with independent control laws applied in underactuated systems |
2022-02-09T12:42:52.553 | <p>I have seen a lot of YouTube videos that use a virtual board for teaching. For example, in the video link below, how can I do the same thing? What are the steps in details in order to be able to create a board as has been done by this professor?</p>
<p><a href="https://www.youtube.com/watch?v=HAPTxiRIZMg" rel="nofollow noreferrer">https://www.youtube.com/watch?v=HAPTxiRIZMg</a></p>
<p>Any help will be very appreciated!</p>
| |computer-engineering|computer-hardware| | <p>This appears to be UV florescent ink, on glass. Have a look at 2min32s when he smudges it with his fingers. In addition, mirroring the video L/R then corrects the fact you are looking at the back of the text, fixing the mirror writing you would otherwise see.</p>
| 49680 | Create a virtual board as often seen on YouTube |
2022-02-09T13:11:06.243 | <p>A 10kg block is suspended from a cable wrapped around a 5kg disk, as shown in the figure.
If the spring has a stiffness constant of k=200N/m, determine the natural period of vibration for the system</p>
<p><a href="https://i.stack.imgur.com/6gsLF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6gsLF.png" alt="enter image description here" /></a></p>
<p>The power energy of the system is given by</p>
<p><a href="https://i.stack.imgur.com/k747N.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k747N.png" alt="enter image description here" /></a></p>
<p>Why is the work produced by the spring positive and the work produced by the weight negative?</p>
<p>Since the spring is opposing the motion and the weight of the block "helps" the motion, I thought the spring did negative work and the weight did positive work.</p>
| |vibration| | <p>When you compress the spring by <span class="math-container">$dx$</span> the energy stored in the spring is given by:</p>
<p><span class="math-container">$$E_s = \frac{1}{2} k \cdot (dx)^2$$</span></p>
<p>By raising to the square power, the energy quantity becomes positive and that means that energy is stored in the system. I.e. however you compress the spring energy is always stored in the system.</p>
<hr />
<p>On the other hand, when you displace a mass by the same amount dx, the energy in the system is not the same in either direction. Going up means that you are storing energy (in this convention positive), while going down means that the potential energy of the system lowers (therefore negative). This is shown by the :</p>
<p><span class="math-container">$$E_p = -m\cdot g \cdot dx$$</span></p>
<p><strong>Note</strong>: the acceleration of gravity in the above formula should have a negative sign (if x is positive upwards).</p>
| 49682 | natural period of vibration for the system - positive and negative work |
2022-02-09T19:16:01.440 | <p>I don't know what the right term for the machine is, but I would like to look into machines that can move something with two bars. For example, the machine that moves the basketball hoop in Stuff Made Here's never-miss basketball hoop. Does anyone know what this is actually called and how I might start to build one that can move, say, six inches in any direction?</p>
| |mechanical-engineering| | <p>Such systems are called Gantry systems. If it is two axis, say X and Y, then it is called and X-Y Gantry system.</p>
| 49687 | How does a double axis movement system work? |
2022-02-11T08:47:09.897 | <p><a href="https://i.stack.imgur.com/B54GR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B54GR.png" alt="enter image description here" /></a></p>
<p>I just used a simple car tire air pump with a gauge like the one in the picture. The device has a handle, a gauge and a button. Squeezing the handle fills the tire with air, and the button releases it.</p>
<p>I noticed, that when the handle is squeezed, the gauge reading momentarily drops to almost zero. When I let go, the reading jumps back up, this time with a higher pressure reading. <strong>How do these gauges work internally, and why does the reading momentarily drop when the tire is filling?</strong></p>
| |pressure|wheels| | <p>The implementation may vary but, to my understanding, the basic idea is that the switch of the handle rotates a plunger (for lack of a better word), which connects the tire to either the pressure gauge or the air supply</p>
<p><a href="https://i.stack.imgur.com/aYzej.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aYzej.png" alt="enter image description here" /></a></p>
<p><strong>figure: a generic schematic - only for illustration purposes.</strong></p>
<p>The handle trigger is spring loaded so it returns to its position.</p>
<p>So what happens when you connect the pressure gauge, is that the tire gets connected with the pressure gauge and you read the pressure.</p>
<p>When you press the button, the rotation of the plunger/valve results into the external air pressure feed to be connected to the tire. This results in increasing the pressure in the tire.</p>
<p>When you release the handle, the plunger returns to its default position and the pressure is properly read again.</p>
<hr />
<p>When the handle trigger is activated, after the momentary drop at the bottom, you can see that the pressure gauge <em>sometimes</em> reads a lower value. This may be due to leakage from the port, or how the ports are connected (again this may vary).</p>
| 49709 | Why does pressure gauge/pump reading drop when the valve is open? |
2022-02-11T15:13:50.480 | <p>In reinforced concrete design, there are some cases that engineers have to design beams which rebar doesn't develop its complete length or hook length according to codes. It is because of restrictions of architecture. The solution from some engineers considers that the connection beam-column in structural analysis is not rigid but free of moments.</p>
<p>Until now, I didn't find research to endorses it. Maybe you can comment about experience or investigations.</p>
<p>The mandatory solution is to increase the column length or decrease de rebar diameter. Is there another solution?</p>
| |structural-analysis|building-design|reinforced-concrete| | <p>If we need a flexible joint having underdeveloped bound in short embedment is not the right way. It is actually a time-bomb waiting to go off and crash the beam explosively.</p>
<p>All the rebar regardless of what is their task have to be fully developed.</p>
<p>If there are needs for a flexible joint there are ways to design a flexible joint, but never an underdeveloped bar. In concrete, by design, failure starts in rebar, not in concrete. If there is a lack of space a bar can bend or form into a hook to develop.</p>
<p>Because of the ductility of the steel, it deflects by a large amount, signaling imminent failure, before the collapse. This gives ample time to occupants for evacuation.</p>
| 49715 | Beam rebar doesn't develop its length in column. Reinforced concrete analysis and design |
2022-02-11T15:31:37.010 | <p>I was watching the video: Equations of Motion for the Multi Degree of Freedom (MDOF) Problem Using LaGrange's Equations from Good Vibrations with Freeball</p>
<p>However, I didn't understand how to obtain these 2 terms (in yellow)</p>
<p>Can someone explain it?</p>
<p><a href="https://i.stack.imgur.com/H4MDj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H4MDj.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/8sAMD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8sAMD.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/7qy4c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7qy4c.png" alt="enter image description here" /></a></p>
| |vibration| | <p>The first one <span class="math-container">$x_2- x_1= 3r\cdot \theta$</span> is very easy. This is based on the fact that for a circle R the arc described by an angle <span class="math-container">$\theta$</span> is equal to <span class="math-container">$R\cdot \theta$</span>. In this case the external radius is 3r.</p>
<p>To understand it easier, it helps me to try and understand intuitively what the equation says for the problem.</p>
<p>So, in this case <span class="math-container">$x_2 - x_1$</span> the length of the rope between the wheel and mass 3m. If you translate the wheel --<em>without rotating it</em>-- then the length of the rope between the wheel and the 3m mass will not change.</p>
<p>Or if you keep O fixed in space and rotate the wheel by <span class="math-container">$\theta$</span>, then the length of the hanging rope will change by <span class="math-container">$3r \theta$</span> because the external radius of the wheel is 3r.</p>
<hr />
<p>Regarding the other term <span class="math-container">$\left(\frac{1}{2} (2k) \left(x_3 -(x_1-r\theta)\right)^2\right)$</span> the term in the parenthesis is the extension of the spring marked as 2k (above mass m). If the quantity <span class="math-container">$x_3 -(x_1-r\theta)$</span> is :</p>
<ul>
<li>positive: the spring marked 2k will be in extension</li>
<li>negative: the spring marked 2k will be in compression</li>
</ul>
<p>In simple cases such as this, I usually do the following thought experiment. I change only one degree of freedom, and see the effect on the element of interest. E.g.</p>
<ul>
<li>if mass m moves downwards (and the wheel does not rotate and its center O does not move) then the spring will be extended by (positive) <span class="math-container">$x_3$</span></li>
<li>if the the center of the wheel O moves downwards by (positive) <span class="math-container">$x_1$</span> (and mass m remains fixed, and wheel does not rotate) then the spring 2k state will be in compression (i.e <span class="math-container">$-x_1$</span>)</li>
<li>if the wheel rotates by an angle <span class="math-container">$\theta$</span> CW (and mass m, and the wheel do not translate), then again the spring will be in extension.</li>
</ul>
| 49716 | Equations of Motion for the Multi Degree of Freedom (MDOF) |
2022-02-11T21:39:04.350 | <p>Please consider an air to air heat exchanger.</p>
<p>On one side, we have 200°C hot air entering the heat exchanger, and leaving it at 150°C.
And on the other side, we have 15°C fresh air to the inlet. Temperature is increased up to 130°C to the outlet.</p>
<p>So, hot air temperature difference is 50°C and cold air temperature difference is 115°C.</p>
<p>Considering this formula:
<span class="math-container">$$\dot{Q} = \dot{m}_{hot}\cdot C_{hot}\cdot (T_{hot1}- T_{hot2}) = \dot{m}_{cold}\cdot C_{cold}\cdot (T_{cold2} - T_{cold1}) $$</span></p>
<p><span class="math-container">$\dot{Q}$</span> = heat exchange rate in W <br />
<span class="math-container">$\dot{m}$</span> = mass flowrate in m/s <br />
C = specific heat <br />
T1 = temperature in<br />
T2 = temperature out</p>
<p>I wonder what would be the resulting temperature of cold air on exit if hot air temperature would be 250°C in and 200°C out. This formula would give me 130°C or so in any case since only the difference between temperature in and temperature out matters.</p>
<p>I think that <em>C</em> depends of the temperature but it seems to not vary that much to make a real difference in the resulting temperature.</p>
| |heat-transfer| | <p>The specific heat of gases does change with temperature. The higher the temperature, the higher the specific heat. But from 200°C to 250°C, there won't be a significant change in its specific heat value. So we can just stick with the formula using nominal specific heat values to find the exit temperature of cold air. Also, for gases, there are two specific heat values; specific heat at constant pressure, <span class="math-container">$C_p$</span>, and specific heat at constant volume, <span class="math-container">$C_v$</span>. Usually, in heat exchangers, we use <span class="math-container">$ C_p $</span> to calculate heat transfer.</p>
<p>You can refer to this link to get an idea of how specific heat changes with temperature: <a href="https://www.ohio.edu/mechanical/thermo/property_tables/air/air_Cp_Cv.html#:%7E:text=The%20nominal%20values%20used%20for,one%20can%20obtain%20significant%20errors." rel="nofollow noreferrer">Specific Heat Capacities of Air</a></p>
<p>Hope this helped.</p>
| 49721 | How important is a hot fluid temperature in a heat exchange? |
2022-02-12T06:25:59.867 | <p>First of all, I apologize for lack of information, but I have to start somewhere.</p>
<p>For those who have knowledge about this topic, can you tell me how PC can not handle more than 8 cameras while DVR/NVR handle many more?
I wonder if DVR/NVR uses special technique or its hardware is designed that way to handle cameras ? Or is it about the cameras that are special for /dedicated to DVRs/NVRs?</p>
| |computer-hardware| | <p>It's going to be difficult to answer your question when the foundation of the question is in error. A search using "linux dvr software" returns many options. I selected the first and did a little digging.</p>
<p><a href="https://www.bluecherrydvr.com/product/v3license/" rel="nofollow noreferrer">Bluecherry</a> is a Linux (PC) based program which allows one to use a computer as a DVR. The license page (linked) references an upper limit of 150 cameras.</p>
<p>I did not dig deeper, but I've seen other Linux DVR systems in the past, many of which are free. One of the free programs I've found is <a href="https://learncctv.com/software-for-multiple-cctv-dvr-works-with-h-264-dvr/" rel="nofollow noreferrer">Universal CMS DVR</a> which references in the setup instructions up to 64 cameras.</p>
<p>With wireless cameras and network connected cameras, one is not limited to USB ports or similar related capacity restrictions.</p>
| 49728 | How can the DVR/NVR handle many cameras while PC can't? |
2022-02-13T05:41:44.927 | <p>I was watching this video on the Navier-Stokes Equation (<a href="https://www.youtube.com/watch?v=XoefjJdFq6k" rel="nofollow noreferrer">https://www.youtube.com/watch?v=XoefjJdFq6k</a>) and had the following questions:</p>
<ul>
<li><p><strong>@ 0:53 Prove that for any given condition, the Navier-Stokes Equation will spit out a solution that will last for all time</strong> : Does this mean that solving the Navier-Stokes Equation will mean that it will become possible to <strong>perfectly predict</strong> how a fluid will behave at any time in the future based on its current condition - or does it mean that this would allow to predict how a fluid will behave with a range of possible behaviors?</p>
</li>
<li><p><strong>@ 4:09 What does it mean for a function to be bigger than another function?</strong> Can someone please elaborate on this?</p>
</li>
<li><p><strong>@ 4:19 A series of intricate inequalities to bound that term by a series of linear terms</strong> : Does this mean that for 2 dimensions, the behavior of a fluid at any future time can be mathematically bounded?</p>
</li>
<li><p><strong>@ 4:33 The argument collapsed in 3 dimensions</strong> : Can someone give an example of a fluid system in 2 dimensions vs. a fluid system in 3 dimensions? What do "dimensions" correspond to in the Navier-Stokes Equation?</p>
</li>
<li><p>@ <strong>4:44 They tried solving an easier version of the equation called the Weak Navier-Stokes Equation</strong> : Is this what is currently being used in real life engineering applications?</p>
</li>
</ul>
<p>In short, I can understand the main idea - we can not prove that the Navier Stokes Equation will provide a solution that is accurate for all future conditions. But at the same time, I think that the Navier-Stokes Equation is extensively being used to analyze complex real world fluid systems (note: I do not have an engineering background). Does this mean that:</p>
<ul>
<li><p>The Navier-Stokes Equation currently is able to well predict the behavior of any fluid at any time in practice, but we can not mathematically prove this fact?</p>
</li>
<li><p>As the video implied, a simplified version of the Navier-Stokes Equation is currently used in complex real world situations - this simplified Navier-Stokes Equation is solvable, but in return does not provide a fully accurate solution?</p>
</li>
</ul>
<p>Can someone please help me understand these ideas behind the Navier-Stokes Equation?</p>
<p>Thanks!</p>
| |fluid-mechanics| | <hr />
<blockquote>
<p>@ 0:53 Prove that for any given condition, the Navier-Stokes Equation will spit out a solution that will last for all time</p>
</blockquote>
<p>No. It just says it can keep predicting something like the right result. No infinities in result, which is already a good achievement. Your bathtub doesnt explode occasionally, water behaves 'normally'. We want that property, some sort of simulation that doesnt produce infinities.</p>
<hr />
<blockquote>
<p>@ 4:09 What does it mean for a function to be bigger than another function? </p>
</blockquote>
<hr />
<blockquote>
<p>@ 4:19 Does this mean that for 2 dimensions, the behavior of a fluid at any future time can be mathematically bounded?</p>
</blockquote>
<p>More like it is restricted to some range of values. Navier stokes is not a perfect simulation, it consist of a lot of averaging. If you restricted conditions are broad enough and there is not much turbulence, then probably.</p>
<hr />
<blockquote>
<p>@ 4:33 The argument collapsed in 3 dimensions : Can someone give an example of a fluid system in 2 dimensions vs. a fluid system in 3 dimensions?</p>
</blockquote>
<p>What do "dimensions" correspond to in the Navier-Stokes Equation? - Meaning is number of directions or gradients around the target. Math is in slightly changed formulas to account for more variables.</p>
<hr />
<blockquote>
<p>@ 4:44 They tried solving an easier version of the equation called the Weak Navier-Stokes Equation : Is this what is currently being used in real life engineering applications?</p>
</blockquote>
<p>Rarely. What used in real life is very far from it. And very specific to a fluid that is being simulated and even to a hardware. GPU can better solve cells, so we use those. For example this <a href="https://en.m.wikipedia.org/wiki/Lattice_Boltzmann_methods" rel="nofollow noreferrer">https://en.m.wikipedia.org/wiki/Lattice_Boltzmann_methods</a></p>
<p>The Navier-Stokes Equation currently is able to well predict the behavior of any fluid at any time in practice, but we can not mathematically prove this fact? - No. It is just an approximation. It is nowhere near close to be able to predict turbulences, and nothing can. It can make similarly behaving simulation, with vortexes of similar size and frequency of appearing, which is good enough usually. Without vortexes it can be solved quite precise.</p>
<p>As the video implied, a simplified version of the Navier-Stokes Equation is currently used in complex real world situations - this simplified Navier-Stokes Equation is solvable, but in return does not provide a fully accurate solution? - Result is better as we improve resolution in time and space and use tricks. We dont use navier stokes directly because other methods in the same time can provide better precision through better time or space resolution or more useful tricks. The more further you go away from a general solution, the more benefits for your particular task you can get.</p>
| 49753 | Understanding the Navier-Stokes Equation |
2022-02-13T23:45:37.543 | <p>I have a heavy object standing on 4 legs (about 30 mm in diameter each) on a wooden floor. The object is soon going to become even more heavier and I am concerned the legs could penetrate the floor. As a precautionary measure I am about to put a metal plate under the legs to spread the load:</p>
<p><a href="https://i.stack.imgur.com/d6dST.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d6dST.png" alt="enter image description here" /></a></p>
<p>Would a steel plate of the same thickness spread the load significantly better than an aluminium one? Which property of the metal is the plate's load-spreading performance a function of — strength (tensile, shear), hardness? Is this function linear? Would a twice-as-thick aluminium plate perform not worse than a steel plate?</p>
| |stresses|metals|strength| | <p>Steel is three times the density of aluminum and about three times everything else too, strength and stiffness modulus included. But stiffness of an actual object scales to the cube of the thickness. So an aluminum plate of the same weight as steel will be 3x thicker but a 9 times stiffer than a steel plate. An aluminum plate twice the thickness of a steel plate will be 2.6x stiffer.</p>
| 49759 | Choosing metal plate to spread point load |
2022-02-14T09:51:06.813 | <p>Generally, for a given loading condition, the bending moment in a beam varies throughout its length due to which the maximum normal/bending stress at every section is different. For a given allowable stress (fixed by the material and Factor of Safety), the maximum bending stress will be maximum where the bending moment is maximum, and will be equal to the allowable. Everywhere else the maximum bending stress will be below this allowable value.</p>
<p>To save material we can vary the section modulus such that the maximum bending stress at every section remains the same, equal to allowable stress. such a beam will have, at every cross section, the same value of max bending stress and hence will be fully stressed.</p>
<p>Such beams are also called beams of uniform strength. <strong>Even though I get why they are called fully stressed beams, I don't understand why they are called beams of uniform strength.</strong></p>
<p>Strength to my understanding is the capability of a structure to bear loads. The larger the loads it can carry the higher is its strength. I don't understand how to use this definition with the concept above.</p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam| | <p>The two types of beams are incompatible by simply observing the stress formula,</p>
<p><span class="math-container">$M = f_bS_x$</span></p>
<p>The name "uniform strength" indicates wherever on the beam, the moment capacity (the strength) is the same/identical, which can be achieved only if we hold both the stress and the section modulus to be constant. However, this defeats the purpose to "<strong>save</strong>" the material by varying the section modulus and requiring the stress to satisfy the unique condition, <span class="math-container">$f_b \le f_a$</span>, that
inevitably will result in varying moment capacity along the span of the beam. So the label "uniform strength" is not applicable. Instead, I think the bam can be said to have "conforming strength" in view of the moment diagram/demand.</p>
| 49766 | Beam of uniform strength |
2022-02-14T17:31:25.850 | <p>I am watching the below video on statically indeterminate axial loading. When he makes a cut on the left half, I understand why he draws the internal force acting to the right, because if he drew it acting to the left then the cut would be accelerating to the left. But when he makes the cut in the right hand side, things are not so simple. He states, "I always draw the internal force as if it were in tension and let the sign take care of itself" even though it is clearly in compression. Here is a screenshot of that part. He is obtaining an expression for PBC.</p>
<p><a href="https://i.stack.imgur.com/rf9zi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rf9zi.png" alt="enter image description here" /></a></p>
<p>Working the rest of the problem he finds reaction A as 600 lb. The problem comes if we had drawn the internal force in the opposite direction. If we had done that and solved as he did we would have found</p>
<p><span class="math-container">$$R_A\cdot (12''-18'') + 18000\cdot lb\cdot in = 0 \rightarrow R_A = \frac{-18000}{-6} = 3000 lb$$</span></p>
<p>So, it is more than just letting the signs take care of themselves. There is a real difference!</p>
<p>In sum:</p>
<p><strong>How do I know what direction to put the internal force for a generic problem?</strong></p>
<p><a href="https://www.youtube.com/watch?v=d3NElXJGwdc" rel="nofollow noreferrer">https://www.youtube.com/watch?v=d3NElXJGwdc</a></p>
| |statics| | <p>(Without knowing the details of your calculations - so I am making pretty heavy inferences), the usual mistake is that the compatibility equation would be different.</p>
<p>In the video the assumption is that the internal forces (both <span class="math-container">$P_{AB}$</span>, and <span class="math-container">$P_{BC}$</span>) have a direction that causes extension. That is why the compatibility equation is:</p>
<p><span class="math-container">$$\delta_{AB} + \delta_{BC} =0 $$</span></p>
<p>I.e. the added extension that both forces cause needs to be zero. This by extension mean that if one of the internal forces is in tension the other will be in compression.</p>
<hr />
<p>if you draw <em><strong>only</strong></em> the internal reaction in the section BC as compressive <span class="math-container">$P_{BC}$</span>, then you need to take account for that into the compatibility equation. I.e.</p>
<p><span class="math-container">$$\delta_{AB} \color{red}{\mathbf{-}}\delta_{BC} =0 $$</span></p>
<p>Then the <span class="math-container">$P_{BC}$</span> inverted formula (i.e. <span class="math-container">$P_{BC}= -P_A + 1000$</span>) will yield the same result.</p>
<hr />
<p><strong>Note</strong> this is about the stressed word <em><strong>only</strong></em> in the previous section:. If both <span class="math-container">$P_{AB}$</span> and <span class="math-container">$P_{BC}$</span> were drawn as compressive, then you'd need to use the <span class="math-container">$-\delta_{AB} - \delta_{BC} =0$</span> (which is identical to <span class="math-container">$\delta_{AB} + \delta_{BC} =0$</span>)</p>
<hr />
<p>Please note that this is only my assumption to what went wrong from extrapolating from your question and from my past experiences explaining to other people. In order to help you more, you'd need to provide more explicit details of your calculation.</p>
| 49770 | How can I justify a consistent sign convention for statically indeterminate problems? |
2022-02-14T20:58:44.677 | <p>Consider a composite beam formed by sandwiching a wood beam between two steel plates, whose cross section is shown.</p>
<p><a href="https://i.stack.imgur.com/7wakr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7wakr.png" alt="enter image description here" /></a></p>
<p>Say the beam is loaded by some arbitrary loading, due to which at any cross section of this composite beam the bending moment is <span class="math-container">$M$</span>. The bending moment <span class="math-container">$M$</span> at that section will be distributed between wood and steel. Let them be <span class="math-container">$M_w$</span> and <span class="math-container">$M_s$</span>, such that</p>
<p><span class="math-container">$$M= M_s + M_w$$</span></p>
<p>is there any way I can determine in what ratio will the bending moment be distributed between wood and steel, i.e.</p>
<p><span class="math-container">$$\frac{M_s}{M_w}$$</span></p>
<hr />
<p>Edit:
I was actually trying to mathematically justify that a composite beam like the one in figure can carry more bending moment than a wooden beam of same cross section and area. I need to get your opinion on its correctness.</p>
<p>Let the allowable stresses in wood and steel be <span class="math-container">$\sigma_{a,w}$</span> and <span class="math-container">$\sigma_{a,s} $</span></p>
<p>The maximum bending moment that wood will be able to carry will be,</p>
<p><span class="math-container">$$(M_{max})_w = \sigma_{a,w} \,Z_w$$</span></p>
<p>where <span class="math-container">$Z_w$</span> is the section modulus of the wood portion.</p>
<p>Similarly,</p>
<p>The maximum bending moment that steel will be able to carry will be,</p>
<p><span class="math-container">$$(M_{max})_s = \sigma_{a,s} \,Z_s$$</span></p>
<p>where <span class="math-container">$Z_s$</span> is the section modulus of the steel portion.</p>
<p>At first I thought I can simply sum these values to obtain the maximum bending moment that the composite beam can carry, but after some thought concluded that can't be the case. So,</p>
<p>If the ratio of the bending moments in each portion at any cross section if the composite beam were to be loaded is,</p>
<p><span class="math-container">$$\frac{M_s}{M_w} = \frac{E_s I_s}{E_w I_w}$$</span></p>
<p>then the maximum beding moment which the composite beam can carry will be,</p>
<p><span class="math-container">$$M_{max,composite} = (M_{max,})_w + \frac{E_s I_s}{E_w I_w}(M_{max,})_w $$</span></p>
<p><span class="math-container">$$M_{max,composite} = (M_{max,})_w (1+ \frac{E_s I_s}{E_w I_w})$$</span></p>
<p><span class="math-container">$$M_{max,composite} = \sigma_{a,w} \,Z_w(1+ \frac{E_s I_s}{E_w I_w}) = M_2$$</span></p>
<p>if the entire beam were to made up of wood then the maximum bending moment that the beam would have carried would be,</p>
<p><span class="math-container">$$M_{max,homogeneous} = \sigma_{a,w} \,Z = M_1$$</span></p>
<p>where <span class="math-container">$Z$</span> is the section modulus of the beam entirely made of wood (having the same cross sectional area and shape as that of composite beam)</p>
<p>Taking the ratio,</p>
<p><span class="math-container">$$\frac{M_2}{M_1} = \frac{Z_w}{Z} (1+ \frac{E_s I_s}{E_w I_w}) $$</span></p>
<p>Am I correct up to this point? If yes, how should I proceed further to show that <span class="math-container">$M_2 > M_1$</span></p>
| |mechanical-engineering|structural-engineering|beam| | <p>You can multiply the width of steel,</p>
<p><span class="math-container">$B$</span>, by</p>
<p><span class="math-container">$n=\frac{E_{steel}}{E_{wood}} $</span></p>
<p>Now the second area moment of the composite beam Ic is the I of the section nBs *h subtracted by the I of the two missing sides on the wood.</p>
<p>'</p>
<p><a href="https://i.stack.imgur.com/5fpJy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5fpJy.png" alt="bm" /></a></p>
<p><span class="math-container">$I_{c} = \frac{ nB *h^3}{ 12} - \frac{B(n-1)*(h-2t)^3}{12}$</span>.</p>
<p>and <span class="math-container">$I_{wood} =\frac{B* (h-2t)^3}{12}$</span></p>
<p>t = Thickness of the steel flange</p>
<p><span class="math-container">$I_{steel} =I_{c}-I{wood}$</span></p>
<p>We can calculate the stress at any point distanced C</p>
<p><span class="math-container">$ \sigma= MC/I_c$</span></p>
<p>except if <span class="math-container">$C>(h-t)/2 $</span> we multiply <span class="math-container">$ \sigma$</span> by n.</p>
| 49777 | Determining the ratio of bending moments in parts of a composite beam |