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2022-02-15T14:06:52.640 | <p>I'm trying to design a controller that will achieve reference tracking of the <em>second</em> state for the following system.</p>
<p><span class="math-container">$$
\begin{aligned}
\dot{x} &=\underbrace{\left[\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right]}_A x+\underbrace{\left[\begin{array}{l}
0 \\
1
\end{array}\right]}_B u \\
y &=\underbrace{\left[\begin{array}{ll}
1 & 0
\end{array}\right]}_C x
\end{aligned}
$$</span>
where
<span class="math-container">$$x=\left[\begin{array}{l}
q \\
\dot{q}
\end{array}\right]
$$</span>
This problem should be equivalent to tracking a ramp on the first state, hence we require a type-2 system if we are to achieve zero steady-state error. Therefore, we require a double integrator to augment the system with the integrals of the error between the reference and performance output. The error <span class="math-container">$e$</span>, reference <span class="math-container">$r$</span> (step) and performance matric <span class="math-container">$H = \left[\begin{array}{l}
0 & 1
\end{array}\right]$</span> are related as
<span class="math-container">$$e=r-Hx$$</span>
If <span class="math-container">$e=0$</span>, the reference is tracked with zero steady-state error. The system can be augmented by adding states <span class="math-container">$\int e$</span> and <span class="math-container">$\int \int e$</span>, with dynamics
<span class="math-container">$$\dot{w}=\underbrace{\left[\begin{array}{l}
0 & 0 \\ 1 & 0
\end{array}\right]}_Mw+\underbrace{\left[\begin{array}{l}
1 \\ 0
\end{array}\right]}_Ne$$</span>
where
<span class="math-container">$$w=\left[\begin{array}{l}
\int e \\ \int \int e
\end{array}\right]$$</span>
The augmented system is then given by
<span class="math-container">$$\left[\begin{array}{l}
\dot{x} \\ \dot{w}
\end{array}\right]=\underbrace{\left[\begin{array}{l}
A & 0_{2 \times 2}\\ -NH & M
\end{array}\right]}_{A_e}\left[\begin{array}{l}
x \\ w
\end{array}\right]+\underbrace{\left[\begin{array}{l}
B \\ 0_{2 \times 1}
\end{array}\right]}_{B_e}u+\left[\begin{array}{l}
0_{2 \times 1} \\ N
\end{array}\right]r$$</span>
<strong>My question:</strong> strangely enough, the pair <span class="math-container">$(A_e,B_e)$</span> is uncontrollable for the provided matrices, but it is controllable when I change to <span class="math-container">$H = \left[\begin{array}{l}
1 & 0
\end{array}\right]$</span>. After designing a controller, it can track a ramp on the state <span class="math-container">$q$</span>, which implies tracking a step on <span class="math-container">$\dot{q}$</span>, but I feel that the former <span class="math-container">$H$</span> should yield identical behaviour when a step <span class="math-container">$r$</span> is applied to track <span class="math-container">$\dot{q}$</span>. Is there an explanation for this?</p>
| |control-engineering|control-theory| | <p>The uncontrollability makes sense, because the first error integral will be a shifted version of <span class="math-container">$q$</span>, hence we cannot control them independently.</p>
| 49787 | Unexpected uncontrollable double integrator system during reference tracking (augmented state-space) |
2022-02-15T14:54:06.963 | <p>The textbook that I'm following for studying bending stresses in beams, states that - "Engineers purposely design beams in this manner (composite manner) in order to develop a more efficient means for supporting loads"</p>
<p>I do get an intuition that for same cross sectional area and shape a composite beam will be able to take more bending moment than a homogeneous beam, but I was trying to prove it mathematically.
Consider for instance, a wooden beam with some given cross sectional area and length-</p>
<p><a href="https://i.stack.imgur.com/FkNcl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FkNcl.jpg" alt="enter image description here" /></a></p>
<p>Let's say the maximum bending moment this beam can carry is <span class="math-container">$M_1$</span></p>
<p>If I were to attach two steel plates at the top and bottom by keeping the height and width the same</p>
<p><a href="https://i.stack.imgur.com/sSdB0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sSdB0.jpg" alt="enter image description here" /></a></p>
<p>then let <span class="math-container">$M_2$</span> be the maximum bending moment that this composite beam can carry</p>
<p>How can I prove <span class="math-container">$M_2 > M_1$</span>?</p>
<p>This question is followed by a question that I asked earlier - <a href="https://engineering.stackexchange.com/questions/49777/determining-the-ratio-of-bending-moments-in-parts-of-a-composite-beam">Determining the ratio of bending moments in parts of a composite beam</a></p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam| | <p>I find it a bit easier to explore this in terms of stiffness.</p>
<p>The strength of a cross-section to bending is</p>
<p><span class="math-container">$$M_R = \dfrac{f_yI}{c} = \dfrac{\epsilon_yEI}{c}$$</span></p>
<p>where <span class="math-container">$f_y$</span> is the yield stress, which is itself equal to <span class="math-container">$\epsilon_yE$</span> (yield strain times elastic modulus); <span class="math-container">$I$</span> is the moment of inertia; and <span class="math-container">$c$</span> is the largest distance to the centroid (half the height, in your case).</p>
<p>Now, we can further simplify this by defining stiffness as <span class="math-container">$K = EI$</span>.</p>
<p><span class="math-container">$$M_R = \dfrac{K\epsilon_y}{c}$$</span></p>
<p>So, the bending moment for a section is linearly proportional to its stiffness <span class="math-container">$K$</span>, its yield strain <span class="math-container">$\epsilon_y$</span>, and inversely proportional to <span class="math-container">$c$</span>. (Obviously, <span class="math-container">$c$</span> and <span class="math-container">$K$</span> are related since both are a function of the section height)</p>
<p>Now let's focus our attention on <span class="math-container">$K$</span>. It is a simple product of two independent variables: <span class="math-container">$E$</span>, which describes the material, and <span class="math-container">$I$</span>, which describes the geometry. However, <span class="math-container">$K$</span> doesn't really differentiate between them: if we have two beams A and B, where A is made of a material that's twice as hard as B, but B's inertia is twice that of A, then they will have the exact same bending strength.</p>
<p>This means we can play around with the actual values of <span class="math-container">$E$</span> and <span class="math-container">$I$</span> that we use in our beam if we want. Specifically, we can use this to transform a composite beam into an equivalent uniform one.</p>
<p>In your case, you've replaced some wooden parts of the beam with steel. You can then change that steel back into "wood" as long as you modify the moment of inertia of those parts accordingly. This is done by "widening" those segments by <span class="math-container">$\dfrac{E_s}{E_w}$</span>. Equivalently, you could pretend the whole beam is steel and then "thin" the wooden segment by <span class="math-container">$\dfrac{E_w}{E_s}$</span>.</p>
<p>That is, a rectangular beam made of wood and steel plates above and below is identical to a wooden I-shape beam, as below (forgive the sloppiest mspaint job of all time):</p>
<p><a href="https://i.stack.imgur.com/C3zPe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C3zPe.png" alt="enter image description here" /></a></p>
| 49789 | Proving the advantage of a composite beam |
2022-02-15T15:08:56.273 | <p>Can flow rate of a small gas flow be controlled by a small solenoid valve that is opened and closed rapidly in a certain frequency?</p>
| |gas|valves| | <p>Yes absolutely. Many small electronic pressure regulators work in pulsed mode. Typical examples:</p>
<p><a href="https://www.hoerbiger.com/en-3/pages/534" rel="nofollow noreferrer">https://www.hoerbiger.com/en-3/pages/534</a></p>
<p><a href="https://content2.smcetech.com/pdf/ITV_4017A.pdf" rel="nofollow noreferrer">https://content2.smcetech.com/pdf/ITV_4017A.pdf</a></p>
<p>A piezo valve would often be used though, rather than a solenoid valve.</p>
<p>I've prototyped a specialty pulsed-mode solenoid-valve regulator, and got outstanding performance. The limitation was durability, which led to the product being changed to proportional-mode solenoid valve.</p>
<p>With the pulsed mode, to maximize control performance, the off-the-shelf solenoid valve that was used, required opening and closing the valves quickly. This in turn had two drawbacks. (1) It excited mechanical resonance (ie overshoot in the mechanical step response) and (2) it put extremely high cycle counts on the moving part. So you need a valve designed for such use.</p>
<p>This is where piezo valves come out ahead at the smallest scales. For slightly larger scales, proportional-mode solenoid valves might be a better fit, although the nice thing about pulsed-mode is that the flow is more predictable, which simplifies control.</p>
<p>In any case, you would need a detailed understanding of the valve choice.</p>
| 49790 | Can flow rate of a small gas flow be controlled by a valve that is opened and closed rapidly in a certain frequency? |
2022-02-16T05:25:03.733 | <p>I was thinking about the realization of a pottery oven, and I wondered which material I could to make the inner walls. Most often the inner walls seem to be made of some kind of hard mineral wool. I also had the idea to make them in fire clay, but then I wondered if titanium would hold such temperatures (up to 1200-1300°C).</p>
<p>It seems mechanically OK, but I feared there might be a possible problem of oxidization, if not even a risk of fire. At what temperature might such kind of problems appear with titanium ?</p>
| |structural-engineering|materials|temperature| | <p>The titanium melting temperature is close to 1670 deg C.
However most titanium alloys have a <a href="https://www.amt-advanced-materials-technology.com/materials/titanium-high-temperature/" rel="nofollow noreferrer">maximum service temperature of 600 deg C</a>.</p>
<p>In any case, purchasing and manufacturing cost would be prohibitive compared to the alternatives that you've mentioned.</p>
| 49802 | Up to what temperature, can titanium be safely used? |
2022-02-16T08:00:34.780 | <p>For school I was asked to model the transfer function <span class="math-container">$H(s)=\frac{3}{(s+4)(s+5)}$</span> in both the time and frequency domains using initial conditions <span class="math-container">$y(0)=2,y'(0)=3$</span>, a step input, and Simulink. I would appreciate any help in better understanding this problem and any real world example would also be appreciated as the coursework is not typically related to real applications.</p>
<p><strong>TL;DR</strong>:</p>
<ol>
<li>How do I pick which factor to apply each initial condition to in the frequency domain (<span class="math-container">$s+4$</span> vs. <span class="math-container">$s+5$</span>)?</li>
<li>What changes should I make in my Simulink models to reach the same system response in both the time and frequency domain?</li>
<li>Do you have any words of wisdom to help me understand this topic more in depth?</li>
</ol>
<p>The differential equation of this system is <span class="math-container">$3(e^{-4t}-e^{-5t})$</span> which I used to set up the frequency domain simulation, summing <span class="math-container">$\frac{3}{s^2-9s-20}$</span> with <span class="math-container">$y(0)*3e^{-4t}$</span> and <span class="math-container">$y'(0)*3e^{-5t}$</span>.</p>
<p><a href="https://i.stack.imgur.com/wv4O6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wv4O6.png" alt="frequency domain model" /></a></p>
<p>My professor claimed the following response was accurate, but I have not been able to get the time domain representation using integrators to match.</p>
<p><a href="https://i.stack.imgur.com/C6Y8E.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C6Y8E.png" alt="frequency domain response" /></a></p>
<p>Here is my attempt at the time domain model:</p>
<p><a href="https://i.stack.imgur.com/jTAzR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jTAzR.png" alt="time domain model" /></a></p>
<p>I passed <span class="math-container">$y'(0)=3$</span> into <code>Integrator</code> and <span class="math-container">$y(0)=2$</span> into <code>Integrator1</code> as initial conditions, which yields the response:</p>
<p><a href="https://i.stack.imgur.com/JRuWl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JRuWl.png" alt="time domain response" /></a></p>
<p>It is my understanding the two responses should be the same. I can get a very similar response in the frequency domain by switching which expression to multiply with the initial conditions (that is swapping the constant blocks used as product inputs) but it is not exact.</p>
| |control-engineering|simulation|matlab|frequency-response|simulink| | <p>For the Simulink model, you can incorporate the initial conditions in the Simulink model. If <span class="math-container">$y(t)$</span> is the output and <span class="math-container">$r(t)$</span> is the input, then the differential equation for the transfer function is:
<span class="math-container">$$\frac{\mathbb{d}^2 y(t)}{\mathbb{d}t} + 9\frac{\mathbb{d}y(t)}{\mathbb{d}t} +20y(t)=3 r(t)$$</span></p>
<p>This means that:
<span class="math-container">$$Y(s)=\frac{(s+9)y(0) + y'(0)}{(s+4)(s+5)}+\frac{3}{(s+4)(s+5)}R(s)$$</span>
Of course, the transfer function <span class="math-container">$H(s)$</span> is obtained when all initial conditions are zero:
<span class="math-container">$$H(s)=\frac{Y(s)}{R(s)}=\frac{3}{(s+4)(s+5)}$$</span>
Therefore you can simply model the following in simulink and use the values for <span class="math-container">$y(0)=2$</span> and <span class="math-container">$y'(0)=3$</span>:
<span class="math-container">$$Y(s)=\frac{2(s+9) + 3}{(s+4)(s+5)}+\frac{3}{(s+4)(s+5)}R(s)$$</span></p>
<p>Now, the time domain solution is:
<span class="math-container">$$y(t)=\frac{49}{4} \mathrm{e}^{-4t}-\frac{52}{5}\mathrm{e}^{-5t}+\frac{3}{20}$$</span></p>
<p>The Simulink diagram is shown below:
<a href="https://i.stack.imgur.com/jTA8B.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jTA8B.png" alt="enter image description here" /></a></p>
<p>The resulting plots are given below:
<a href="https://i.stack.imgur.com/DTcJz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DTcJz.png" alt="enter image description here" /></a></p>
<p>You can see that both solutions are identical.</p>
<p>Note: In modelling the initial conditions, we want the system response when the input <span class="math-container">$r(t)$</span> is zero. This is the zero input response which can be achieved by using an impulse response to the homogenous solution. And an impulse response can be generated by differentiating a step response. Hence the <span class="math-container">$s$</span> in the transfer functions that represent the initial conditions.</p>
| 49805 | Modeling a second order system in time and frequency domain |
2022-02-16T19:14:49.333 | <p>I'm planning on using a lead screw for my robotic arm project. My question was that if I for example use a 3Nm torque motor and couple it to a lead screw (four-start lead screw), does the torque remain the same? Or does it increase?</p>
<p>How do I calculate the torque that the lead screw will give out? I know that a lead screw converts rotary motion into linear motion.</p>
<p><a href="https://i.stack.imgur.com/DZT2H.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DZT2H.jpg" alt="Lead Screw Driven Robot Arm" /></a></p>
<p>Thank you in advance for any help</p>
| |torque| | <p>It is all just leverage, mechanical advantage. You can solve all of it through ratio of movements on each side.</p>
<p>Assuming 3 N/m and screw advances about 8 mm per 1 turn and its radius is 4 mm.</p>
<p>Forst find linear force, tangential at screw's side in this case</p>
<p>3 N/m / 0.004 m</p>
<p>Gives 750 N at the screw's side, tangential. To get axial force, find ratio of movement:</p>
<p>750 / ( 0.004 * 2 * 3.14) * 0.008</p>
<p>Gives 239 N axial force of the screw. Then you also draw another lever at your arm, but it ratio of shoulders is close to 1, so it can be ignored. If it is far from 1:1, calculate it too. If hand part is 3 times as long as the muscle part, then you get 80 N at the hand part.</p>
<p>And yes, worm gear or screw drive give large mechanical advantage that is used for making slow and strong movements.</p>
| 49808 | How does a lead screw affect the torque of a motor? Can a leadscrew be used as way of increasing torque? |
2022-02-17T09:47:56.070 | <p><a href="https://i.stack.imgur.com/rvbYm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rvbYm.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/MPP4X.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MPP4X.jpg" alt="enter image description here" /></a></p>
<p>This is an example problem from the book vector mechanics for engineers. I had a doubt that how the position vectors for A and position vector B relative to A defined. That is the equation which is rounded in yellow in the second picture, there we should write the vertical component of of B/A and add it to position vector A. But we are adding magnitudes of lengths of vectors to obtain the length of wire. It is confusing. Please help</p>
| |mechanical-engineering|applied-mechanics| | <p>In general you are right. It's not possible to <em>casually</em> add position vectors in different coordinate systems.</p>
<p>However in this case it is possible, because of the context. In order to help you understand it more intuitively, it might help you are considering that the constrain equation describes the <em>length of the rope</em>. And the values <span class="math-container">$x_A$</span> and <span class="math-container">$x_{B|A}$</span> are not vectors but they are <strong>scalar coordinates of the position vectors</strong>. The fact that the coordinate systems are selected in such a way that the rope is parallel, means that it is possible to use those coordinates directly in the constraint equation.</p>
| 49816 | Does the constraint equations written for pulley mass system can be any coordinate system other than rectangular |
2022-02-17T16:20:37.420 | <p>I saw this flood light pole. It has two LED bulbs, but I do not know what the box in the middle is (it was not an LED), could you guys help me understand what this could be? This flood light is located in a plaza parking lot called Churchill square, in St John's, NL, Canada.</p>
<p><a href="https://i.stack.imgur.com/XOl2U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XOl2U.png" alt="enter image description here" /></a></p>
| |electrical-engineering|power-electronics|electrical|lighting| | <p><a href="https://m.youtube.com/watch?v=1VSkhz8y-hY" rel="nofollow noreferrer">https://m.youtube.com/watch?v=1VSkhz8y-hY</a></p>
<p>In here you can see, at 17th second, that these poles are located at the corners of the parking lot of the shopping center. And they are not present elsewhere in the city.</p>
<p>It is not a camera, those needs round casing.</p>
<p>It is not a network tower, they are too close and those are looking like vertical box and dont like other boxes nearby. But lower frequencies can work fine with LEDs nearby.</p>
<p>It is not a PV, its a rather bulky box.</p>
<p>It is not a LED as you say.</p>
<p>I dont think it is a power supppy unit, as those are cheaper to service at a lower height.</p>
<p>My guess is that it is some sort of digital fence system for the shopping center that operates at a much lower frequency than a phone.</p>
<p>Probably to track trolleys as they are moved away and to lock their wheel to prevent people taking these trolleys elsewhere.</p>
<p>Tower height is excessive, and light lamps are added at a lower height. Meaning these towers need their height for something else. Digital fence benefits from such a height, as it doesnt get obstructed by vehicles, and provides more even power of a sigal within its zone.</p>
| 49820 | Flood light pole components |
2022-02-17T16:48:26.297 | <p><a href="https://i.stack.imgur.com/vGtzU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vGtzU.png" alt="Diagram" /></a></p>
<p>Torque due to force of gravity:<br />
-Payload: 0.2kg x 0.3m x 9.81 = 0.5886<br />
-Hand: 0.1kg x 0.25m x 9.81 = 0.24525<br />
-Forearm: 0.3kg x 0.1m x 9.81 = 0.2943</p>
<p>Torque due to force of gravity = 1.13Nm</p>
<p>What about torque due to angular acceleration?<br />
I know it is Torque = Moment of inertia x angular acceleration,
However, the motor will be rotating the hand and the forearm around the x-axis.</p>
<p>So, how would I calculate the torque for this motion?</p>
| |motors|torque| | <p>I assume you are trying to measure the torque needed to turn the arm horizontally to right or left.</p>
<p>If that's the case we calculate the angular moment, I as:</p>
<p><span class="math-container">$$ I= \Sigma m_i r^2_i \ = \\ 0.2kg*0.3^2m+0.1kg*0.25^2m+0.3kg*0.1^2m \\=0.18+0.00625+0.003=0.027Nm^2$$</span></p>
<p>Not that we don't have g factor, and torque is:</p>
<p><span class="math-container">$$T=\alpha*I$$</span></p>
| 49821 | How would I calculate the torque of the motor shown in the diagram? |
2022-02-18T00:29:26.193 | <p>I'm confused about stepper motor speed. On their data sheet it doesn't really specify a maximum speed. I want to use a stepper motor with a 3:1 gear ratio. This is the motor that I am looking at:</p>
<p><a href="https://www.amazon.co.uk/RTELLIGENT-Stepper-425oz-4-lead-57x57x100mm/dp/B07V9RMWMK/ref=sr_1_6?crid=3C9Y1WFVZD1S0&keywords=nema+23&qid=1645121856&sprefix=nema+23%2Caps%2C143&sr=8-6" rel="nofollow noreferrer">https://www.amazon.co.uk/RTELLIGENT-Stepper-425oz-4-lead-57x57x100mm/dp/B07V9RMWMK/ref=sr_1_6?crid=3C9Y1WFVZD1S0&keywords=nema+23&qid=1645121856&sprefix=nema+23%2Caps%2C143&sr=8-6</a></p>
<p>So with the gear ratio, the torque will increase from 3Nm to 9Nm, and the speed will be reduced by 3 times.</p>
<p>However, what will the speed be?</p>
<p><a href="https://i.stack.imgur.com/wU97H.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wU97H.png" alt="enter image description here" /></a></p>
<p>I appreciate any help on helping me understand.</p>
| |motors|gears|torque|stepper-motor| | <p>If a gear is used on the shaft of the motor, then assuming:</p>
<ul>
<li><span class="math-container">$T_M$</span> is the motor torque (also the gear input)</li>
<li><span class="math-container">$n_M$</span> is the rpm (also the gear input)</li>
<li><span class="math-container">$T_g$</span> is the gear output torque</li>
<li><span class="math-container">$n_g$</span> is the gear output rpm</li>
<li><span class="math-container">$i$</span> is the gear ratio (assume that <span class="math-container">$i>1$</span> means a reduction of rpm)</li>
</ul>
<p>then the following equation hold:
<span class="math-container">$$T_g = i\cdot T_M, \qquad n_g = \frac{n_M}{i}$$</span></p>
<p>So there is a tradeoff.</p>
<hr />
<h2>Torque (a bit more complex)</h2>
<p>Torque of a stepper motor is a bit more complex. I.e. the relationship between torque and rpm is not linear, but in reality is more similar to the image below.</p>
<p><a href="https://i.stack.imgur.com/pHxft.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pHxft.png" alt="enter image description here" /></a></p>
<p>So, with increasing rpm on the motor the Torque output is affected. So you might need to consult the datasheet of your motor manufacturer.</p>
<p>Additionally, other factors like the selected driver of the stepper motor can have an effect depending on the supplied voltage.</p>
| 49827 | Could I have help on understanding stepper motor speed? |
2022-02-18T15:21:38.103 | <p>I'm planning on using this servo for my robotic arm project:</p>
<p><a href="https://www.amazon.co.uk/yusvwkj-Digital-Torque-Waterproof-Control/dp/B08MTB1X84/ref=sr_1_2_sspa?crid=3FW87FTTM5LPT&keywords=35kg+servo&qid=1645197363&sprefix=35kg+servo%2Caps%2C101&sr=8-2-spons&psc=1&spLa=ZW5jcnlwdGVkUXVhbGlmaWVyPUEzTVhNMjIwQ09UM1BWJmVuY3J5cHRlZElkPUEwNzI2MzM0MktPU0pBOU9PRU1FRCZlbmNyeXB0ZWRBZElkPUEwMTQ0MDU2M1JOV0JKM0NRRVFMWCZ3aWRnZXROYW1lPXNwX2F0ZiZhY3Rpb249Y2xpY2tSZWRpcmVjdCZkb05vdExvZ0NsaWNrPXRydWU=" rel="nofollow noreferrer">Amazon link</a>.</p>
<p>It's a digital servo with 35 kg-cm (3.4 Nm) torque. However, when looking at other robotic arm projects, I've mainly seen analog servos being used. Is there a reason for that?</p>
<p><strong>Will I be able to use this servo for my robot arm project with my Arduino and a suitable motor drive?</strong></p>
| |motors|robotics|servo| | <p>Analogue and digital servos work the same way from the user's point of view. Internally, the demanded position and the measured position are compared, the difference being called the error. The calculation of the error and the subsequent control of the motor can be done with analogue or digital electronics. The latter is meant to offer a more intelligent control of the motor. This may improve the response time and allow the servo to be more sensitive to small errors without overshooting.</p>
| 49835 | Digital servo for my robot arm? |
2022-02-19T03:08:56.017 | <p>When choosing a power supply for a stepper motor, e.g. this one below:</p>
<p><a href="https://www.amazon.co.uk/RTELLIGENT-Stepper-425oz-4-lead-57x57x100mm/dp/B07V9RMWMK/ref=mp_s_a_1_3?crid=3NN5NC2HYQOQS&keywords=rtelligent%20nema%2023&qid=1645122809&sprefix=rtelligent%20nema%2023%2Caps%2C48&sr=8-3" rel="nofollow noreferrer">Amazon Link NEMA 23</a></p>
<p>This one has:
Rated Current: 4.0A
Operating Voltage: 24-50VDC</p>
<p>The Stepper Driver I will be using is:</p>
<p><a href="https://www.amazon.co.uk/Digital-Microstep-Stepper-Controller-2-phase/dp/B07YWZRXGR/ref=mp_s_a_1_11?crid=WUX47R3BQI2I&keywords=dm542&qid=1645210783&sprefix=dm542%2Caps%2C52&sr=8-11" rel="nofollow noreferrer">Stepper Driver</a></p>
<p>Voltage Range:20-50VDC
Current Range: 1-4.2A</p>
<p>So would my power supply need a output Current of 4.0A?</p>
<p>And a Output Voltage of 24-50VDC, how is the performance of the motor affected if I use 24VDC vs 50VDC Power Supply?</p>
<p>Thank you in advance!</p>
| |motors|torque|stepper-motor|specifications| | <p>For static and low-speed applications, it doesn't matter. If you need full torque at speed, it's nice to have a datasheet that provides a graph like this:</p>
<p><a href="https://i.stack.imgur.com/IkcXz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IkcXz.png" alt="torque-speed-voltage" /></a>
<em>image from: nanotec.com; note semilog scale</em></p>
<p>Roughly speaking, the effect of increasing the supply voltage from 24V to 48V would roughly double the RPM which is the "corner" on that graph.</p>
<hr />
<p>The main reason is winding inductance. Magnetization is controlled by current, and rate-of-change of current is limited by max voltage. At the high end of the speed range, the lower supply voltage cannot increase the current fast enough.</p>
<p>For example, let's say you have a 200step/rev motor, 4A per winding, 2mH per winding, and 24V supply.</p>
<p>Max dV/dt = V/L = 24/.002 = 12000 A/s.</p>
<p>A very crude approximation of max rate needed is a triangular current profile (in reality the profile will be sinusoidal for effective high speed motion, but let's neglect it here for simplicity). This triangular profile has the current swinging from +4A to -4A and back to +4A, at each step = 8A swing per half step = 16A/step.</p>
<p>Thus the roughly highest rate at which a triangular profile is possible, is (12000 A/s)/(16A/s) = 750 steps/s = 3.75rev/s = 225RPM. This a crudely reckoned limit of where torque would start to fall off.</p>
<p>In real-world applications, at higher speeds it's also good to have excess torque capacity beyond what's "used" by the load, to let the system accelerate/decelerate and cope with resonances. Driver details would play a role there.</p>
| 49839 | How to select the right power supply for a stepper motor? |
2022-02-19T12:57:02.217 | <p>Should the teeth of meshing gears be even rather than odd numbers? If not, which is better?</p>
| |mechanical-engineering|gears|mechanisms|machine-design|machine-elements| | <p>The recommendation is quite the opposite.</p>
<p>The teeth of meshing gears are, where possible, chosen to be odd or, better again, primes so that a bad tooth doesn't keep hitting the same point on the opposing gear and gear wear will be even. For example, 23 tooth gear driven by a 19 tooth gear will only come back into phase after 19 × 23 = revolutions.</p>
<p>You need to confirm this, but as far as I know the tooth repeat frequency is given by
<span class="math-container">$\frac {LCM}{T_1 \times T_2}$</span> where <span class="math-container">$ LCM $</span> is the least common multiple of the number of teeth on the two gears and <span class="math-container">$ T_1 $</span> and <span class="math-container">$ T_2 $</span> are the number of teeth on each gear.</p>
| 49846 | Is it essential for two meshing gears to have even number of teeth? |
2022-02-19T14:21:07.493 | <p>I'm developing a board game and it needs a mechanism of incremental rotation in which the rotated object will snap to one of five possible angles. I have been trying to search for examples of mechanisms that snap in increments when rotated but I don't seem to know the correct terminology.</p>
<p>The rotating shaft would be a five sided polygon (pentagon) as in the picture:</p>
<p><a href="https://i.stack.imgur.com/Zcpv2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zcpv2.jpg" alt="Design" /></a></p>
<p>What kind of design might work in this case so that after the player rotates the shaft, it always snaps into a "straight" angle so that one of the five surfaces of the polygon is facing straight up?</p>
| |mechanisms|mechanical| | <p>A vee-shaped block spung up against the underside of the pentagonal shaft would do the trick.</p>
<p><a href="https://i.stack.imgur.com/qojQd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qojQd.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. A badly dimensioned snap.</em></p>
<p>You'd need a lot longer on the "spring" section of this to allow enough flexibility.</p>
<p>A search for <a href="https://www.google.com/search?q=3d+printed+detent&tbm=isch" rel="nofollow noreferrer">3D printed detent</a> should give plenty of ideas.</p>
| 49849 | Incremental rotation that snaps in place |
2022-02-19T16:29:59.907 | <blockquote>
<p>Two different incompressible fluids a and b flow into a chamber through two circular ports of area 0.1 and 0.2 m<sup>2</sup> respectively.</p>
<ul>
<li>Fluid a is of density 1,200 kgm<sup>−3</sup> and flows directly into the chamber at a speed of 2 ms<sup>−1</sup>.</li>
<li>Fluid b has a density of 1,500 kgm<sup>−3</sup> and has a velocity normal to the surface of the chamber inlet port surface of 3 ms<sup>−1</sup>.</li>
<li>The chamber has a single outlet port of area 0.3 m<sup>2</sup>.</li>
</ul>
</blockquote>
<blockquote>
<p>What is the density and the speed of the mixture coming out of the outlet?</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/tJh7f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tJh7f.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/bllNQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bllNQ.png" alt="enter image description here" /></a></p>
<p>The answer in the back of the textbook says the correct answer is 2.6 m/s for the velocity and 1461 kg/m<sup>3</sup> for the density. Does anyone know why my numbers are slightly off, is it possibly due to the fact that fluid B enters normal to the surface. I have attached the question below.</p>
<p>Thanks in advance for any help</p>
| |fluid-mechanics|homework| | <p>The answer rounded the exit velocity to 2.6 instead of 2.67, then, again, dropped the fraction (0.54) after the decimal point:</p>
<p><span class="math-container">$\gamma_c = \dfrac{(1200*0.1*2)+(1500*0.2*3)}{(0.3*2.6)} = 1461(.54)$</span></p>
| 49854 | Continuity / conservation of mass |
2022-02-20T19:42:52.557 | <p><em>The Question is from shear stresses in beams</em></p>
<p>Consider a beam made of circular cross section, in which at any cross section the shear force is V.</p>
<p><a href="https://i.stack.imgur.com/dVlOR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dVlOR.png" alt="enter image description here" /></a></p>
<p>The textbook I'm following states that -</p>
<blockquote>
<p>When a beam has a circular cross section, we can no longer
assume that the shear stresses act parallel to the y axis. For
instance, we can easily prove that at point m (on the boundary of the
cross section) the shear stress <span class="math-container">$\tau$</span> must act tangent to the boundary.
This observation follows from the fact that the outer surface of the
beam is free of stress, and therefore the shear stress acting on the
cross section can have no component in the radial direction.</p>
</blockquote>
<p><strong>I'm having trouble with understanding why the shear stress at m, should be tangent to the boundary.</strong></p>
<p>Say I enlarge the element,</p>
<p><a href="https://i.stack.imgur.com/pCr5c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pCr5c.jpg" alt="enter image description here" /></a></p>
<p>and assume that the shear stress acts along y axis. this shear stress can be resolved into tangential and radial components to get <span class="math-container">$\tau_t$</span> and <span class="math-container">$\tau_r$</span>. The book says this <span class="math-container">$\tau_r$</span> can't be balanced by any other stress in the radial direction and hence it must be zero. But it could happen that a normal stress is developed on the surface as shown in (b), which balances <span class="math-container">$\tau_r$</span></p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam| | <p>Let's take a small element of the circular shaft and apply the shear stress <span class="math-container">$\tau_1$</span> in an arbitrary direction on the surface of the cut. The shear stress can be resolved into a radial shear stress <span class="math-container">$\tau_r$</span> and tangential shear stress <span class="math-container">$\tau_t$</span> as shown below.</p>
<p><a href="https://i.stack.imgur.com/PgO8x.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PgO8x.png" alt="enter image description here" /></a></p>
<p>We recall the state of shear stresses on a cube element - when shear stress presents on one plane, there must exist a shear stress on the face normal to that plane. In this case, there must be a shear stress <span class="math-container">$\tau_z$</span> corresponding to <span class="math-container">$\tau_r$</span>, however, we know that the stress on the outermost surface of an element is zero, from this realization we can conclude that <span class="math-container">$\tau_r$</span> must be zero as well, so only <span class="math-container">$\tau_t$</span> remains effective.</p>
<p>Note that this phenomenon is valid for shear stress on the perimeter only.</p>
| 49879 | Direction of Shear Stress on the periphery of a circular section beam |
2022-02-21T18:09:23.510 | <p>Sometimes people say "The engine of Car A has a higher torque than Car B with both having the same horsepower" My question is : if both cars are equal in horsepower can't we just <strong>avail</strong> this equal horsepower by gearsets and get equal torques for both cars by reducing speeds for the two cars?And thus a truck and a car will have equal torques just by gear sets "If" they have the same horsepower . Why is all that struggle about The car Torque number when the key is the horsepower that can give any speed or torque by having gearsets?</p>
| |mechanical-engineering|gears|torque|car|engines| | <p>Although I agree with others in that the reason in not technical (but rather advertising - I guess the suffix <em>power</em> in horsepower resonates well with hormone driven masculinity), I will try to explain some of the subtle differences I perceive.</p>
<p>Below is a sample engine performance curve. With green/blue solid line is the torque, while with dashed red line is the horsepower. On the x axis is the rpm.</p>
<p><a href="https://i.stack.imgur.com/UlQKD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UlQKD.png" alt="enter image description here" /></a></p>
<p>Immediately. is is evident that horsepower tends to change more than the torque. This is a reflected in the equation of :</p>
<p><span class="math-container">$$P = M \cdot \omega$$</span></p>
<p>where:</p>
<ul>
<li>P is the horsepower</li>
<li>M is the torque</li>
<li><span class="math-container">$\omega$</span> is the angular velocity (which is proportional to the rpm <span class="math-container">$\omega = {2\pi \cdot n\over 60}$</span></li>
</ul>
<p>Having a flat torque curve is considered a good characteristic. That would yield an almost straight horsepower curve. In both cases (horsepower and torque) the value that is reported is the maximum value of the curve.</p>
<p>E.g.: for torque in the example above the torque would be 230Nm at 3700 rpm, and the power would be 120 PS at 5500 rpm.</p>
<p>I agree that the most useful number for everyday driving would be the torque, because it is at a range that most cars usually work. The only case that max power is reached is when accelerating hard and hitting over 5000 rpm in which case the engine is stressed and probably borderline abused. So torque would be a more useful everyday metric.</p>
<hr />
<p>However, in general , maximum horsepower and maximum torque give only a snapshot of the engine performance at different rpms. If you really need to understand your engine and you need to optimise the driving attitude, then you'd need to have access to the full mapping of the torque/power curve.</p>
<p>For example ,things can become more complicated because engines can be tweaked to yield different torque responses (see below).</p>
<p><a href="https://i.stack.imgur.com/XdJam.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XdJam.jpg" alt="enter image description here" /></a>
<strong>figure source :<a href="https://www.carthrottle.com/post/66dq846/" rel="nofollow noreferrer">Car Throttle</a></strong></p>
<p>All those engines above have the same peak horsepower, but completely different torques. Driving those cars would give completely different sensations. It is interesting to note that the top speed would be the same, however accelerating would be different.</p>
| 49892 | Question about torque in car engines |
2022-02-22T10:04:18.977 | <p>Consider a cantilever beam with a point load at its free end. If we cut the beam through a plane as shown (in which the load lies), to divide the beam into two parts, and if we isolate individual parts, then will there be any stresses on the plane that I point out in the figure?
<a href="https://i.stack.imgur.com/2oQkk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2oQkk.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering|civil-engineering|beam| | <p>Using Euler's beam theory we don't have transverse stresses.</p>
<p>The cantilever example deforms under the load and deflects but the cross-section remains square (the original shape and size).</p>
<p>But we know (and ignore) that by Hooks law The top of the beam narrows due to the elongation under deformation and the bottom thickens.</p>
<p>So the cross-section will look like a trapezoid, not a square. This transversal deformation increases as the radius of beam deflection decreases (more curvature) near the support. And these strains cause transversal stresses on the cut section you investigate.</p>
<p>On top, we have transversal tension perpendicular to the surface of your half-cut and on the bottom compression stresses.</p>
| 49902 | Are there any stresses on this plane in a beam? |
2022-02-22T17:27:06.780 | <p>I was wondering whether it was possible to apply Bernoulli's equation to determine what pressure your hand would experience if you were to hold it out the window of a driving car? If so, how could you use Bernoulli's equation to solve the problem?</p>
| |fluid-mechanics| | <p>I believe that you could, in a very basic sense, answer this question using Bernoulli's equation. Assuming no loss from the boundary layer of the vehicle, analyzing at a point in the center of the hand at which we assume the velocity of the air is approaching zero relative to the car.</p>
<p>P<sub>1</sub> + ½ (ρ) v<sub>1</sub><sup>2</sup> = P<sub>2</sub> + ½ (ρ) v<sub>2</sub><sup>2</sup></p>
<p>Ignoring the ρgh portions since they will cancel out.</p>
<p>Point 1 being a place adjacent to the vehicle traveling at (let's say) 88 ft/s.</p>
<p>Point 2 being a point at the center of the hand where the air speed is being forced to approach zero. ½ (ρ) v<sub>2</sub><sup>2</sup> term then is estimated to be zero.</p>
<p>Now:
P<sub>2</sub> = P<sub>1</sub> + ½ (ρ) v<sub>1</sub><sup>2</sup></p>
<p>Assuming atmospheric pressure at 14.7 psi.
Assuming density of air at 0.080713 lb/cubic foot.
Assuming the vehicle travels at 88 ft/s.</p>
<p>½ × (0.080713) × 88<sup>2</sup> = 312.5 psf</p>
<p>312.5 psf / 144 = 2.17 psi</p>
<p>14.7 psi + 2.17 psi = 16.87 psi (or 2.17 additional psi)</p>
<p>Thoughts? That's just my first pass at it.</p>
| 49916 | Bernouli equation |
2022-02-23T06:07:27.167 | <p>I am working on a piece of art that has parts that rotate slowly, at about 1RPM, thanks to a couple of bearings. There is a possibility that it might be installed beside a beach for about a week. The problem is that sand and bearings do not play together well.</p>
<p>Or do they? Is there a bearing type that can resist the ingress of sand, or failing that, a reasonably reliable and durable way of preventing sand entering?</p>
<p>Yes, I am considering covering it with boots, but this question is part of research to see if there is an alternative.</p>
| |bearings| | <p>Bering's by name bear all of the friction at a point of two or more surfaces grinding together. They will all wear out. Your use case is "by a beach for a few days". Just suck it up and save ur money. They will work just fine. The art piece is not meant to last if it has moving parts.</p>
| 49923 | Is there such a thing as a sand-proof bearing? |
2022-02-23T10:12:26.587 | <p><a href="https://i.stack.imgur.com/Uoszp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Uoszp.png" alt="enter image description here" /></a></p>
<p>What does this kind of welding symbol indicate? I can see that it is a field weld, all around, with flat finish but the two vertical lines are not known to me. I found it in a drawing of a steel railing. <strong>What kind of weld is this, and how is it made?</strong></p>
| |civil-engineering|welding| | <p>This is square groove all around site weld joint. The circle at the arrow and reference line intersection means all around. The flag means Field Weld joint. 2 vertical lines means a square groove. The final weld finish is flush shown by horizontal line placed on the weld symbol.
This is a helpful article to <a href="https://www.materialwelding.com/welding-symbols-explained-by-chart-and-drawing/" rel="nofollow noreferrer">learn Welding Symbols</a> in depth.</p>
| 49928 | What is this kind of weld? |
2022-02-23T11:50:52.453 | <p>I'm working on a project where part of a unit needs to rotate around a fixed point, and be supported on two fixed (i.e. not castor) wheels.</p>
<p>When experimenting with the wheels, it feels like fixing them at 90 degrees to the unit results in them 'scraping' across the floor somewhat, as the arc of travel is always slightly misaligned with the plane of their rotation. Further, it feels like holding them at a slight camber mitigates or even eliminates this effect. However, I have no way of confirming this.</p>
<p>Is this basic observation correct, and if so, is there a method for calculating the optimal camber (a and b) for wheels different distances (d and e) from the pivot point x?</p>
<p><em>(Note, if it helps/matters, the wheels have a more rounded profile than those shown in the diagram.)</em></p>
<p><a href="https://i.stack.imgur.com/gQc1i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gQc1i.png" alt="simple diagram of system" /></a></p>
| |wheels| | <p>The optimum angle would be that of a pair of cones sharing the pivot point and each cone's axis running through the axis of its respective wheel.</p>
<p><a href="https://i.stack.imgur.com/3hKXN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3hKXN.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. The correct camber can be determined by aligning each wheel's axis with the point of rotation.</em></p>
<p>The calculation is basic trignometry.</p>
<p><a href="https://i.stack.imgur.com/6sYNA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6sYNA.png" alt="enter image description here" /></a></p>
<p><em>Figure 2. These conic rollers have the same end diameters. They will roll perfectly around the shared point but the dark one will have to rotate at a faster rate.</em></p>
| 49930 | Is there an optimal camber for fixed wheels rotating around a point? |
2022-02-23T20:56:14.010 | <p>So, I was just doing some Finite Element Analysis (FEA) to figure out what would be the effect on the stiffness of a cantilever beam if another beam is attached to it on its free end.</p>
<p>Below shows the two cases I am using as examples.</p>
<p><a href="https://i.stack.imgur.com/Rk0Qv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rk0Qv.png" alt="enter image description here" /></a></p>
<p>I am only conducting geometric linear analysis i.e. change of stiffness of the green beam is not tracked by the solver during loading. Here are the deformation results for this beam.</p>
<p><a href="https://i.stack.imgur.com/PHgmQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PHgmQ.png" alt="enter image description here" /></a></p>
<p>So apparently, the deformation along Z axis only changes by a magnitude of 0.018 mm of the green cantilever beam. But my question is that is this change in deformation because of the addition of the vertical gray beam on the free end of it, or its just a result of the numerical errors coming from the FEA solver's nature?</p>
<p>I mean if the deformation is changing for the green body, then it means that its stiffness has changed since the solver is linear and same load is applied. So attaching another body, like the gray body in this example, somewhere along the length of the green body should change its stiffness (and therefore its resulting deformation) or not? And how would you anticipate the affect of increasing the height or area of connecting face of the gray body to green body would affect the green body's stiffness and its deformations?</p>
<p>If I talk about purely Statics, then the bodies are assumed to be Rigid. If I talk about Mechanics of Materials, then bodies are assumed to have very small deflections and rotations. Plus, the beam thoeries like Bernoulli and Timoshenko assume that the cross section is constant throughout the length of the beam. But what is the reality for this example I have mentioned here in my post?</p>
| |beam|deformation|stiffness| | <p>The integration equation for the deflection of a cantilever beam provides the answer for the small difference between the two models.</p>
<p><span class="math-container">$\delta_y = \iint \frac{M}{EI}dx + Ax +B$</span></p>
<p>Note, For beams with varying cross-sections, you need to break up the integration accordingly. And, the result will change if you add member self-weight in the analysis.</p>
<p><strong>Comment:</strong></p>
<p>The strain is greatest at the fixed end is an unchangeable fact. We know well that for a cantilever beam, the maximum moment and normal stresses occur at the support, thus the displacements (lengthen/shorten) of the extreme fibers. The moment and stresses then decrease from the maximum at the support to zero at the free end, as well as the strain.</p>
| 49943 | Should the stiffness of this cantilever beam be different for these two conditions? |
2022-02-25T15:13:30.740 | <p>Consider a homogeneous beam with a rectangular c/s, its neutral axis will be at equal distances from the top and bottom edges.</p>
<p><a href="https://i.stack.imgur.com/VYySY.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VYySY.jpg" alt="enter image description here" /></a></p>
<p>The first moment of the entire area about the NA will be zero, since we have as many <span class="math-container">$y\,dA$</span> terms as <span class="math-container">$-y\,dA$</span> terms.</p>
<p>I was investigating whether this is true for other c/s shapes too, which are not doubly symmetric, for example,</p>
<p><a href="https://i.stack.imgur.com/P1qEX.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P1qEX.jpg" alt="enter image description here" /></a></p>
<p>Will the first moment of area <span class="math-container">$\int_A y\,dA$</span> about the NA be equal to zero for all c/s shapes? Even for the one that I show right above?</p>
| |mechanical-engineering|structural-engineering|stresses|beam| | <p>Yes, the sum of the first moment of inertia will be zero. That is actually how we find the neutral axis.</p>
<p>Let's call the are on top of the neutral axis area T and the bellow axis area B.</p>
<p>Then the location of the neutral axis with respect to a line on the bottom of the C/S will be.
<span class="math-container">$$ \overline{y} = \frac{areaT*\overline{y_t}+areaB* \overline{y_b}}{areaT+areaB}$$</span></p>
<p>It will be closer to the larger end sort of like seeking the line that if you put the C/S on it it will balance itself.</p>
<p>he is an example of a T beam.</p>
<p><a href="https://i.stack.imgur.com/3Ehuk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3Ehuk.png" alt="neutral x" /></a></p>
<p>.</p>
| 49966 | First moment of area about the neutral axis |
2022-02-27T03:25:45.923 | <p><a href="https://i.stack.imgur.com/cSTzP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cSTzP.jpg" alt="enter image description here" /></a></p>
<p><strong>TLDR: Is this only a 2:1 pulley system?</strong></p>
<p>Hi there, I’m trying to build a Murphy bed for my roommate. Long story short we’re using an electric hoist + pulley system to raise and lower the bed.</p>
<p>The setup is that the motor is behind the base of the bed on the left. It has a 440 lb pulling capacity. I was planning on running it up the top of the frame, down diagonally where two pulleys would be on each corner of the bed, the wire running from one side of the bed to the other, back up diagonally, up to another pulley on the other side of the top of the frame, then all the way down and anchored to the wall/ground.</p>
<p>The bed is ~330 lbs (~1468 N), so I was doing an analysis of the system to try to see what the pulley ratio would be.</p>
<p>My first assumption is that the two pulleys at the bottom of the bed act as one pulley in practice, as they don’t seem to provide any mechanical advantage. So I simplified my model into a 3 pulley system, but it seems like the weight of the bed is only split on that middle pulley. The other two pulleys don’t seem to provide any mechanical advantage either, they just seem to redirect the force.</p>
<p>Is this analysis correct? Is this only a 2:1 pulley system?</p>
| |mechanical-engineering|pulleys| | <p>Note that the top right pulley does not move so it is doing nothing and may be removed. The right hand rope would then be attached to the wall at the top.</p>
<p>However, I think your drawing shows a support under the end of the bed. This is required because if someone sits on one corner the pulley arrangement alone allows the bed to twist. An improvement would be to keep the two top pulleys and use two ropes each tied to the base of the bed and both wound up by the motor. Alternately, legs that drop out and support the end would solve that problem.</p>
<p><a href="https://i.stack.imgur.com/5rr7o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5rr7o.png" alt="enter image description here" /></a></p>
<p><em>Figure 1. Force vectors.</em></p>
<p>The tension in the diagonal ropes in your second analysis is <span class="math-container">$\frac {734}{cos(A)}$</span> where <em><strong>A</strong></em> is the angle from the vertical. The vertical component will be 734 N on each side and there will be opposing horizontal forces so the rope tension will be higher than 734 N. You'll need to take this into account when specifying the rope, pulleys and anchoring to the wall.</p>
<p>As an aside note that the tension in the rope is proportional to $\frac 1 {cos(A)}. As <em><strong>A</strong></em> approaches 90° the tension goes to infinity. A tightrope requires huge tension compared to a slack rope.</p>
| 49983 | Would this pulley system only halve the force? |
2022-02-28T07:34:52.967 | <p>Adhesive is applied to "soft" ballistic fabrics (like kevlar or even <a href="https://www.youtube.com/watch?v=_7jiIQOgwtI" rel="nofollow noreferrer">fiberglass</a>) in order to decrease <a href="https://www.youtube.com/watch?v=EOfMgXlY3zA" rel="nofollow noreferrer">bulge</a> when being hit by a projectile.</p>
<p>What would be the effect of gluing layers of ballistic fabrics together (effectively, making it stiff), with a powerful adhesive, on the fabric's ability to resist penetration?
Obviously, the adhesive itself will also resist the penetration, but that is not what I ask.</p>
<p>EDIT: Conecring Solar Mike comment about inertia, on <a href="https://www.howitworksdaily.com/question-of-the-day-how-does-kevlar-stop-a-bullet/#:%7E:text=It%20is%20a%20light%2C%20polyarylamide,to%20be%20hard%20to%20stretch." rel="nofollow noreferrer">howitworksdaily</a> it said that "[kevlar] has a high tensile strength. This means it takes a huge amount of energy to make its fibres stretch even a little ... most will absorb the energy of the bullet by stretching a small amount."</p>
<p>So, does the stiffness from gluing will weaken the kevlar by not allowing it to strech, or it will actually make it stronger, because now it's harder to make them strech?</p>
<p>Example of kevlar properties from <a href="https://www.fibermaxcomposites.com/shop/datasheets/aramidFiberMDS.pdf" rel="nofollow noreferrer">Fibermax</a>:</p>
<p>Yarn Type 1111<br />
Linear Density (nominal) [dtex] 1260<br />
f 1000<br />
Linear Density (effective) [dtex] 1350<br />
Finish Content [%] 0.7<br />
Elongation at Break [%] 3.10<br />
Breaking Tenacity [MPa] 2987<br />
Breaking Strength [N] 280<br />
Modulus [GPa] 91<br />
Tube Size [mm] 290x94<br />
Number of Bobbins x Weight per Bobbin [kg] 6 x 4.5 / 90 x 4.5</p>
| |materials|adhesive| | <p>Fabric based ballistic projection works by distributing load of it's force, kinetic energy perpendicular to it's direction.</p>
<p>Adhesives are <a href="https://www.masterbond.com/applications/adhesive-compounds-bonding-kevlar-substrates" rel="nofollow noreferrer">already in use</a> for bonding such fibers.
Adhesives act to prevent the thread weaves from being undone, like poking a needle thru a shirt, often the hole displaces the fiber weave
<a href="https://i.stack.imgur.com/315jw.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/315jw.gif" alt="enter image description here" /></a></p>
| 50002 | Effect of adhesive on ballistic fabrics |
2022-03-01T09:21:59.417 | <p>All quadropole mass spectrometers that I have seen provide a total pressure reading in addition to partial pressure readings for the selected masses.</p>
<p>The mechanism of the total pressure measurement is not described anywhere I have looked.</p>
<p>Does the mass spectrometer use its quadropole and faraday cup or is there simply a hot cathode vacuum gauge somewhere inside?</p>
| |measurements|vacuum|spectrometry| | <p>Looking for this answer in the context of RGAs (rest gas analyzers), I was able to find an answer:</p>
<p><strong>Leybold answers the question in an <a href="https://www.inficon.com/v1/attachment/Transpector2-TechNote-PartialP-TotalP" rel="nofollow noreferrer">application note</a></strong></p>
<p>The total pressure is measured by a <a href="https://en.wikipedia.org/wiki/Hot-filament_ionization_gauge" rel="nofollow noreferrer">hot filament ionization gauge</a>, where the cathode is the filament of the mass spectrometer and the anode is created by an aperture in front of the entry to the quadropole mass filter.</p>
<p>A "total pressure plate" is placed in front of the entry to the quadropole section of the mass spectrometer. Ions that are used for partial pressure measurements pass through a hole in the plate into the quadropole section of the mass spectrometer.
Another portion of the ions collide with the plate. The ion current from the plate can be measured and is correlated to the total pressure.</p>
| 50023 | How does the total pressure detector in a quadropole mass spectrometer work? |
2022-03-03T09:52:21.290 | <p>I want to calculate the velocity of a fluid required to turn a water wheel at a specific angular velocity. The fluid enters the housing of the water wheel through a inlet channel. It turns the wheel and then disappears through the outlet. See the figure for a sketch of the problem. <a href="https://i.stack.imgur.com/Yt9Zw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Yt9Zw.png" alt="Fluid enters a housing at velocity <span class="math-container">$v_{in}$</span>, turning the wheel and leaving at velocity <span class="math-container">$v_{out}$</span>" /></a></p>
<p>Fluid enters at a known velocity <span class="math-container">$v_{in}$</span> through an inlet with known dimensions. The fluid will hit the wheel, causing it to rotate from the impact at an angular velocity of <span class="math-container">$\omega$</span> rad/s. Remaining fluid leaves with a velocity <span class="math-container">$v_{out}$</span>. Due to friction at the axis of the wheel, the moment required to initiate rotation of the wheel is <span class="math-container">$T$</span> Nm. I want to calculate the minimal velocity <span class="math-container">$v_{in}$</span> required to overcome the friction and rotate the wheel and I have found two different approaches.</p>
<p><strong>Approach 1</strong><br />
Using the kinetic energy equation and perseverance of energy gives:</p>
<p><span class="math-container">$\frac{m{v_{in}}^2}{2} = \frac{I{\omega}^2}{2} +\frac{m{v_{out}}^2}{2} + T\theta \tag{1}$</span></p>
<p>where <span class="math-container">$I$</span> is the inertia of the wheel, <span class="math-container">$m$</span> the mass of the fluid control volume and <span class="math-container">$\theta$</span> the angle rotated by the wheel. <span class="math-container">$T\theta$</span> is the work done by the friction at the axis. This approach is quite straight forward, if I know the moment <span class="math-container">$T$</span> and the inlet velocity <span class="math-container">$v_{in}$</span> the maximum <span class="math-container">$\omega$</span> can be calculated or if I want to know <span class="math-container">$v_{out}$</span> for a specific <span class="math-container">$\omega$</span>, that can be computed as well.</p>
<p><strong>Approach 2</strong><br />
Using another form of the energy equation, namely the one for steady incompressible flow, i.e., <em>the extended Bernoulli equation</em>:</p>
<p><span class="math-container">$\frac{p_{out}}{\rho} + \frac{{v_{out}}^2}{2} + gz_{out} = \frac{p_{in}}{\rho} +\frac{{v_{in}}^2}{2} + gz_{in} + w_{shaft} - loss \tag{2}$</span></p>
<p>Where <span class="math-container">$\rho$</span> is the density of the fluid, <span class="math-container">$p$</span> the pressure, <span class="math-container">$z$</span> the height, <span class="math-container">$w_{shaft}$</span> the work done on the shaft and the <span class="math-container">$loss$</span> is from the loss due to friction in the pipe. Dividing with the acceleration of gravity <span class="math-container">$g$</span>, we get the <em>energy per unit weight</em> or the <em>head</em>.</p>
<p><span class="math-container">$\frac{p_{out}}{\gamma} + \frac{{v_{out}}^2}{2g} + z_{out} = \frac{p_{in}}{\gamma} +\frac{{v_{in}}^2}{2g} + z_{in} + h_{s} - h_L \tag{3}$</span></p>
<p>Here the definition of specific wieght is used <span class="math-container">$\gamma = g\rho$</span>. The head loss due to the friction in the pipe is neglected, since the distance traveled is considered short. The head in the shaft can be expressed through work in the shaft as:</p>
<p><span class="math-container">$h_s = w_{shaft}/g = \frac{\dot{W}_{shaft}}{\dot{m}g} = \frac{T\omega}{Q\gamma} \tag{4}$</span></p>
<p>Where <span class="math-container">$\dot{W}$</span> is the time rate of work <span class="math-container">$\dot{W} = T\omega$</span> and <span class="math-container">$Q$</span> is the flow rate, related to the mass flow as <span class="math-container">$Q = {\rho}\dot{m}$</span>. Inserting (4) in (3) with <span class="math-container">$h_L = 0$</span> and since the fluid does work on the wheel, <span class="math-container">$h_s$</span> is negative, energy is taken out from the system. This gives:</p>
<p><span class="math-container">$\frac{p_{out}}{\gamma} + \frac{{v_{out}}^2}{2g} + z_{out} = \frac{p_{in}}{\gamma} +\frac{{v_{in}}^2}{2g} + z_{in} - \frac{T\omega}{Q\gamma} \tag{5}$</span></p>
<p>If there is no pressure drop across the wheel, <span class="math-container">${\Delta}p = 0$</span>, there is no height difference since everything happens in the same plane <span class="math-container">${\Delta}z = 0$</span>, resulting in:</p>
<p><span class="math-container">$\frac{{v_{out}}^2}{2g} = \frac{{v_{in}}^2}{2g} - \frac{T\omega}{Q\gamma} \tag{6}$</span></p>
<p>From equation 6 <span class="math-container">$v_{out}$</span> can be computed if <span class="math-container">$v_{in}$</span>, <span class="math-container">$\omega$</span> and <span class="math-container">$T$</span> are known. Since <span class="math-container">$Q=v_{in}/{A_{in}}$</span> where <span class="math-container">$A_{in}$</span> is the area of the inlet, it is considered to be known once <span class="math-container">$v_{in}$</span> is known. Similar to equation (1), the same parameters can be computed for certain known conditions.</p>
<p>Using these two equations, one might presume that you will get similar results. However, I do not and I wonder why. Is it something which I neglect using one of the equations which is not neglected in the other? Am I missing something? I want to know what is wrong and why I get different results.</p>
<p><strong>Example</strong><br />
Using (6) to find <span class="math-container">$v_{out}$</span> for <span class="math-container">$v_{in} = 5$</span> m/s yields <span class="math-container">$v_{out} = 0.95$</span> m/s. Solving for the same input vales but using (1) insteaed, yileds an imaginary solutions, since the square root becomes negative. <span class="math-container">$v_{out}=\sqrt{v_{in} - \frac{I{\omega}^2}{m} - \frac{2T{\theta}}{m}}$</span> Thus according to equation (1), a higher inlet velocity is required to overcome the resistance at the wheel. Which one is correct?</p>
| |mechanical-engineering|fluid-mechanics|turbines| | <p>I think I found the answer. The Bernouilli equation (6) is for <strong>steady</strong> flow, the flow keeps on coming. Thus it will calculate the velocity required to turn the wheel at a certain angular velocity and torque. Since the flow is steady the wheel will turn at this velocity as long as there is a constant flow of fluid entering the camber.</p>
<p>The energy equation is a bit different and it can be noticed by calculating <span class="math-container">$\theta$</span> in (1) from <span class="math-container">$\omega$</span>, considering that <span class="math-container">$\omega$</span> is known prior. Thus we get <span class="math-container">${\theta} = \int_{0}^{t} \omega dt = {\omega}t+A$</span>, where A is an integration constant. <span class="math-container">$\theta$</span> is the distance traveled by the wheel, at <span class="math-container">$t=0$</span>, that distance is 0, which gives: <span class="math-container">${\theta}(0) ={\omega}{\cdot}0 +A =0$</span>, <span class="math-container">$A = 0$</span>. Thus we get
<span class="math-container">${\theta}(t) = {\omega}t\tag{7}$</span></p>
<p>Inserting (7) in (1), it is easier to see how the energy equation (1) is time dependent, or at least the part associated with the friction is. I think it can be compared to when you kick a ball along a surface. When you kick the ball it has an initial velocity, but because it rolls on a surface, some portion of the initial kinetic energy will be lost to the friction. And that is a function of time, the longer or further you want the ball to roll, the larger initial velocity is required.</p>
<p>In conclusion it could be said that if you have a steady flow through a wheel, the Bernouilli's equation (6) is more suited. However, if you have "package" of fluid which is launched towards the wheel, then the energy equation (1) is suited. E.g., if I want the wheel to turn 3 revolutions around its axis, and I have a finite volume of fluid which I will launch towards it, how fast must the fluid travel.</p>
| 50054 | Fluid dynamics - Velocity required to turn wheel |
2022-03-04T15:56:15.373 | <p>I am working on setting up a system that senses when an object goes though a door. I've seen things using lasers or induction. I have a set of objects smaller than 1 square inch. I just want to know which object goes through the "door". Think of the object as empty and the sensors can go anywhere or everywhere around the door.</p>
<p>The time doesn't need to be too accurate, within a second.</p>
<p>I just haven't decided on a bunch of stuff, I just know that the item should be able to held on your person and the "door" could be a door or just an entranceway to another room ( you know, like a big doorway but without the door). I want to track when someone goes through a "doorway", I don't know how to describe that. The height will be door height and the width could be larger, up to 10 ft I guess. The speed would be walking speed usually. The object would be on the person without being annoying to carry. In my head it's just an object, not sure what else. I would like it to be able to be in a pocket though.</p>
<p>basically I would like to know how many people enter a room and who they are.</p>
| |sensors|electrical|wireless-communication| | <p><a href="https://en.wikipedia.org/wiki/Radio-frequency_identification" rel="nofollow noreferrer">Radio frequency ID</a> <a href="https://en.wikipedia.org/wiki/RF_module" rel="nofollow noreferrer">chips</a> are an easy way to achieve what you want.</p>
<p>Some stores place them on their high valued merchandise items to prevent theft. RF reading devices are located at the entrances and when a RF ID chip, that hasn't been canceled, passes near such a device and alarm is activated.</p>
<p>Some airlines use them to track luggage for the frequent flyer customers.</p>
<p>Such devices can be placed in lanyard that people were around their necks and as they pass through a doorway which has a reader next to it the RF ID chip is noted and a computerized database is updated.</p>
<p>Depending on the application, various sized FR ID devices can be obtained.</p>
| 50065 | how would I identify when and who enter through a doorway, assuming they can have part of the technology on their person |
2022-03-05T14:54:51.733 | <p>Here is a practise question that I am working on. The beam has different thickness due to the supports in the middle. I would expect the maximum bending stress be away from the middle due to the support? But that is not what I got.</p>
<p><a href="https://i.stack.imgur.com/y6Fr8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/y6Fr8.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/16OQs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/16OQs.png" alt="enter image description here" /></a></p>
<p>Also I am very confuse with the moving pointload. Do we have different induced maximum external moment (due to pointload) at different location along the beam(due to the thickness)? For example, the maximum moment at the thinner beam would be <span class="math-container">$(-1m \times 20kN)$</span>? But the maximum moment at the thicker beam would be <span class="math-container">$(-2m \times 20kN)$</span>? (As the pointload moves)</p>
| |structural-engineering|structural-analysis|structures| | <p>What you have is a box beam with a different second moment of inertia at the segment without or with the cover plate, <span class="math-container">$I_1, I_2$</span>, respectively. Note that the varying stiffness does not affect the distribution of shear force and moment along the span; so without guessing, you can compare the resulting bending stresses obtained from 1) place the concentrated loads at the mid-span, and 2) place the concentrated load at where the section change occurs (point "a").</p>
<p><a href="https://i.stack.imgur.com/TdPkr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TdPkr.png" alt="enter image description here" /></a></p>
<p><strong>ADD:</strong> A cross-section is said to be symmetrical when the mirror image exists on the centroidal axis. In the graph below, the cross-section on the upperrow is said to be "doubly symmetrical", while the cross-sections in the lower row are said to be "singularity symmetrical about y-axis", for which, the equation for <span class="math-container">$I_x$</span> in the sketch is not applicable without modification.</p>
<p><a href="https://i.stack.imgur.com/mfTOR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mfTOR.png" alt="enter image description here" /></a></p>
| 50080 | How do you calculate the maximum bending stress in a beam with different thickness along different parts on the beam? |
2022-03-05T20:26:48.777 | <p>The system represented in the figure consists of a 2 kg pulley to which two springs and a rigid square with 10 kg/m are connected, which in turn is articulated at point B (System is in the xz plane).</p>
<p><a href="https://i.stack.imgur.com/aeQ23.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aeQ23.png" alt="enter image description here" /></a></p>
<p>Dynamic equilibrium equations:</p>
<p><a href="https://i.stack.imgur.com/TAIb2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TAIb2.png" alt="enter image description here" /></a></p>
<p>Can someone explain these 2 terms in yellow? It's making me a bit of confusion since I thought it should be "+" and not "-" and both terms should be equal to each other and not symmetrical</p>
<p><a href="https://i.stack.imgur.com/fDyFF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fDyFF.png" alt="enter image description here" /></a></p>
<p>this was the free body diagramm that i draw</p>
<p><a href="https://i.stack.imgur.com/W2Ycc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W2Ycc.png" alt="enter image description here" /></a></p>
<p>In other exercice that i done, they consider JTeta1 and JTeta2 in the opposite direction</p>
<p><a href="https://i.stack.imgur.com/TS4gJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TS4gJ.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/0QHZh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0QHZh.png" alt="enter image description here" /></a></p>
| |vibration| | <ol>
<li>The total term which includes the highlighted portion represents the torque exerted by the vertical spring.</li>
<li>The spring torque depends on spring force which in turn depends on the spring extension / compression.</li>
<li>If both ends of the spring move by equal amount, the spring doesn't get more compressed or extended and hence no change in the spring force. So the <em>difference of the movement</em> between the two ends of the spring determines the extension / compression of the spring.</li>
<li>The highlighted term gives the extension / compression of the spring.
The following diagram shows the movement of the each side of the spring.
The movement of the top end of the spring is <span class="math-container">$r_2 \theta_1$</span> positive in the downward direction due to the choice of the direction of positive <span class="math-container">$\theta_1$</span>.
The movement of the bottom end of the spring is <span class="math-container">$2L \theta_2$</span> which is also positive in the downward direction due to the choice of the direction of positive <span class="math-container">$\theta_2$</span>. Hence the extension / compression of the spring is given by <span class="math-container">$\mp(r_2 \theta_1 - 2L\theta_2)$</span>.</li>
<li>The term is the same in both equations (except for an overall sign as seen in your example) since the extension / compression of the spring is not dependent on which equation we are considering at any instant.</li>
</ol>
<p><a href="https://i.stack.imgur.com/5iH5w.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5iH5w.png" alt="pulley system" /></a></p>
| 50084 | Dynamic equilibrium equations |
2022-03-06T09:34:29.670 | <p>I am currently checking out an appartment in which I noticed a strange connection between the side walls and the ceiling.</p>
<p>The guy told me it is a shadow gap which I then googled and it just didn't seem to fit what I found, at least the way it is constructed/oriented. It looks as they made the walls thicker but not all the way to the ceiling.</p>
<p>I attached a file how it looks, the picture is not to scale. It is a tiny gap where the ceiling and the side wall meet.</p>
<p>The building is a one story concrete building, and the apparmant is on the first floor( also the last floor) so the buliding roof is on top of it.</p>
<p>Is this a normal thing in building? Did any one of you encountered/made this before?</p>
<p>Thank you very much<img src="https://i.stack.imgur.com/gSAK7.png" alt="enter image description here" /></p>
<p>Edit: added a picture<a href="https://i.stack.imgur.com/71aL1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/71aL1.jpg" alt="enter image description here" /></a></p>
<p>Edit2: Higher quality image<a href="https://i.stack.imgur.com/lyisC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lyisC.jpg" alt="enter image description here" /></a></p>
| |structural-engineering|civil-engineering|building-design| | <p>Your sketch indicates a short roof overhang (exterior) or a ceiling decorating block/trim (interior). For the latter case, if it is located a short distance from the ceiling, then it can be called a "shadow gap", which is made to create a more interesting visual effect. At times, it also has the function to absorb sound by using sound-absorbing (foam) panels.</p>
| 50085 | Is this a normal shadow gap? Connection between a wall and the ceiling |
2022-03-07T03:58:52.680 | <p>Consider a fluid flowing over a flat plate with a free stream velocity of <span class="math-container">$U_\infty$</span>. We know that a thin boundary layer region will be formed. The shear stress acting at any distance y (measured perpendicular to the plate) is given by Newton's law of viscosity as,</p>
<p><span class="math-container">$$\tau_y = \mu \frac{\partial u }{\partial y}$$</span></p>
<p>where <span class="math-container">$u$</span> represents velocity which is a function of <span class="math-container">$x$</span> and <span class="math-container">$y$</span></p>
<p><a href="https://i.stack.imgur.com/IRaFx.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IRaFx.jpg" alt="enter image description here" /></a></p>
<p>I was interested in knowing how can I show the stresses acting on a fluid element taken at y. For instance, we take a stress element in <em>mechanics of materials</em>, and show shear stresses with complementary ones, acting on it. However in case of a fluid, an infinitesimal fluid particle is not at rest, it's moving so I'm not sure if complimentary stresses will appear.</p>
<p>In a nutshell, I wanted to know how the stresses on a fluid element will look like, if I took one in the boundary layer region. If somebody could help.</p>
<p><a href="https://i.stack.imgur.com/dmahu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dmahu.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|fluid-mechanics|civil-engineering| | <p>The complementary stresses are still needed: without them, for a differentially small, rectangular-parallelepipedal fluid element, the net torque on the fluid element would contain three differentially-small factors of side lengths of the element, whereas the moment of inertia of the fluid element would contain five differentially-small factors of side lengths of the element. Hence, in the absence of complementary stresses, the reciprocal of the angular acceleration would be differentially small, i.e. the angular acceleration would be infinite.</p>
| 50097 | A fluid element taken in the velocity boundary layer region |
2022-03-07T15:55:09.667 | <p><a href="https://en.wikipedia.org/wiki/Transom_(nautical)" rel="nofollow noreferrer">Transom</a> basically is a flat aft, usually vertical. It reinforces the structure, adds deck area, and because that the flat part is above the waterline, no fluid dynamic is hurt.</p>
<p>The extra deck area is of course utilized for profit in container ship designs:
<a href="https://i.stack.imgur.com/BC721.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BC721.jpg" alt="Stern of the Chinese ship the"Y M Orchid", a 275-meters long recently built cargo ship. Port of Genoa, Italy." /></a>
Image courtesy of International Labour Organization, under Attribution-NonCommercial-NoDerivs 2.0 Generic license.</p>
<p>In navies, there are aircraft carriers/amphibious assault ships/whatever ship equipped with a through flight deck, with the bow flared to support rectangular deck outboard.</p>
<p>There are also <a href="http://rosslillistonewoodenboat.blogspot.com/2012/02/discussion-about-transom-bows-whimbrel.html" rel="nofollow noreferrer">prototypes</a> of yachts with almost vertical transom bows.</p>
<p>My question is: <em>Why don't other ships have a transom bow?</em>
With both directions' transoms, the whole main deck shall become a rectangle, so that more containers can be loaded.</p>
| |marine-engineering|ships|naval-engineering| | <p>The problem is known as slamming.</p>
<p>From Ship Design And Construction - Lamb (2003):</p>
<blockquote>
<p>The problem of slamming, that is when the forward bottom of ships strikes the sea when sailing with low forward draft in rough head seas, causing severe bottom damage, is more of a problem for tankers and bulk carriers due to their
fullness forward (large block coefficients).</p>
</blockquote>
<blockquote>
<p>Damage by wave impact can also happen in the aftbody if formed in an extreme way, for example, flat, horizontal areas above propeller and rudder, and large flat transom sterns.</p>
</blockquote>
<p>As a ship heads directly into waves, it rides up and then slams down into the sea. It is an existing problem for ships with a high block coefficient, <span class="math-container">$C_B$</span>, like bulk carriers if they are unloaded and floating high. Unloaded they travel with full ballast for stability and minimize slamming. Slamming can be mitigated by not travelling directly into waves.</p>
<p>Basic Ship Theory - Rawson et al (2001):</p>
<blockquote>
<p>The block coefficient, <span class="math-container">$C_B$</span>, is the ratio of the volume of displacement to the volume of a rectangular block whose sides are equal to the breadth extreme, the mean draught and the length between perpendiculars.</p>
</blockquote>
<p>The closer it is to 1 the more rectangular the hull.</p>
<p>Above the waterline, you are proposing a transom or flat bow, with a conventional bow below the waterline. An overall wide ship with a high <span class="math-container">$C_B$</span>. Works on a flat ocean, but heading into any waves the wider area above the waterline will experience slamming sending vibration throughout the ship compromizing every joint. Ships are made of steel, but not over engineered.</p>
<p>Even aircraft carriers, which are wide at the flight deck have a flared bow to cut through the waves.</p>
| 50103 | Why aren't there many ships with transom bows? |
2022-03-08T14:52:03.357 | <p>What is the appropriate position to place an accelerometer, along the length of a cantilever beam, for the follwing two cases:</p>
<ol>
<li>The beam is under sinusoidal excitation at the free end.</li>
<li>The beam experience an impact load at the free end.</li>
</ol>
| |measurements|vibration| | <p>Put it at the free end. What I would do here is called a "bump test" it is when you measure the response frequency after an impact using a data collector. I agree with Nmech. You could just hit it and measure the response to get the natural frequency.
You can get the natural frequency from anywhere on the beam in reality, but the amplitudes will likely be largest on the free end.</p>
| 50115 | Positioning accelerometer on a cantilever beam |
2022-03-09T20:23:41.607 | <p>If I have a fan with these attributes</p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th>Category</th>
<th>Value</th>
</tr>
</thead>
<tbody>
<tr>
<td>Air flow</td>
<td>108.2 CFM (3.03 <span class="math-container">$m^3/min$</span>)</td>
</tr>
<tr>
<td>RPM</td>
<td>3100</td>
</tr>
<tr>
<td>Dimensions LxHxW (mm)</td>
<td>120x120x25</td>
</tr>
</tbody>
</table>
</div>
<p>Is it correct to calculate the thrust force at max rpm like this:</p>
<p>Fan air flow = 108.2 CFM = 0.0511 <span class="math-container">$m^3/s$</span></p>
<p>Air density at room temperature = 1.2 <span class="math-container">$kg/m^3$</span></p>
<p>Fan speed = 3100 rpm = 19.48 <span class="math-container">$m/s$</span> given the radius is <span class="math-container">$\frac{\frac{120}{2}}{1000} = 0.06$</span>m</p>
<p><span class="math-container">$F_{thrust}$</span> = Fan air flow * Air density * Fan speed = 0.0511 * 1.2 * 19.48 = 1.19N</p>
| |fluid-mechanics|airflow| | <p>If I remember correctly thrust of a stream of gas is:</p>
<p><span class="math-container">$$F=\dot m *v + (P_e-P_0)A_e$$</span></p>
<p>Assuming the pressure before and after the fan is equal to 1atm:</p>
<p><span class="math-container">$$F= \dot m *v$$</span></p>
<p>So you need to find out the stream's flow speed not the fan speed. (The rotation of the fan is not directly related to the flow speed. One needs to calculate the blade's angle of attack and aspect ratio and even then it will be an approximation.)</p>
<p>Therefore</p>
<p><span class="math-container">$$V= \dot Q/A=\frac{0.051}{\pi 0.06^2}$$</span></p>
| 50128 | Thrust force of a fan |
2022-03-10T15:28:10.787 | <p>I would like to determine the pressure drop of fluid in cross flow over a bank of tubes. (see figure 1).</p>
<p><a href="https://i.stack.imgur.com/XTqo8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XTqo8.png" alt="Horizontal cylinders represent tubes. Arrows parallel to the cylinders are labeled "internal flow of fluid through tube." Arrows perpendicular to tubes are labeled "fluid in cross flow over tube bank."" /></a></p>
<p>Previously, I've used the correlation listed in "Fundamentals of Heat and Mass Transfer" by Incopera and DeWitt. However, this correlation requires looking up coefficients on a graph (see figure 2 - f is used to mean Eu/k by Incopera).</p>
<p><a href="https://i.stack.imgur.com/xdF2V.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xdF2V.png" alt="A graph showing a variety of lines correlating f to Re at a variety of values of Pt, and another graph correlating chi to Pt/Pl at a variety of values of Re" /></a></p>
<p>Now, I'd like calculate a pressure drops across a large number of tube banks in a computer program I'm running, so I need a correlation that's either just one equation or based on a lookup table. I tried using tables 1 and 2 from <a href="https://www.thermopedia.com/content/1211/" rel="nofollow noreferrer">the Thermopedia article on tube banks in crossflow</a>. While this gives reasonable answers at some points, it also produces clearly incorrect answers at others. For example, for a square in-line bank, this correlation shows that Euler's number for b=2.0 exceeds Euler's number for b=1.5 as Re approaches 1e6, which doesn't line up with the graphs and also is obviously an incorrect answer (See figure 3, b=1.5 is in orange and b=2.0 is in green).</p>
<p><a href="https://i.stack.imgur.com/MTWOC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MTWOC.png" alt="Log/log graph showing a blue (b=1.25), orange (b=1.5), and green (b=2.0) line." /></a></p>
<p>Is there a correlation which I could use in my program which is more accurate than the one on Thermopedia?</p>
| |mechanical-engineering|fluid-mechanics|heat-transfer|heat-exchanger| | <p>I developed a solution to this problem and published it as a Python library called <a href="https://github.com/ellieLitwack/tubebank" rel="nofollow noreferrer">tubebank</a>. You can install it using pip. I used the method proposed here by Solar Mike.</p>
<hr />
<p><strong>Edit</strong>
The comment made by Solar Mike, regarding the methodology to solve the problem.</p>
<p><em>So make a data table of x & y values then use vlookup() or interpolation to get the value of y for a given x based on the curve you want. Did this for many characteristics when i first got into excel. There is a program called DataThief which will take an image of a graph and give you x & y values...</em></p>
| 50140 | Method for computing the pressure drop across a bank of tubes |
2022-03-10T17:09:51.413 | <p>The system represented in the figure is composed of two rigid assemblies articulated in
two points and connected to several springs as indicated. Each beam segment weighs 2 kgf/m, and the 5 kgf weight is concentrated at the end F. The stiffness constant k=1000N/m and the length L=1m . The system is operating in the xz plane.
Determine, justifying the expressions used, the frequencies
and natural modes of the system</p>
<p>Determine, justifying the expressions used, the frequencies
and natural modes of the system</p>
<p><a href="https://i.stack.imgur.com/LfLsh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LfLsh.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Nt67l.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nt67l.png" alt="enter image description here" /></a></p>
<p>I'm wondering why "<span class="math-container">$\theta_A$</span>" and "<span class="math-container">$J_A \theta_A$</span>" are in the opposite direction. Can someone explain it?</p>
<p>In the solutions they rotate in this direction</p>
<p><a href="https://i.stack.imgur.com/tNCPX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tNCPX.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/KhsYW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KhsYW.png" alt="enter image description here" /></a></p>
| |vibration| | <p>I think the sketch below explains why they are in the opposite direction. Note that the direction of <span class="math-container">$J_A \theta_A$</span> indicates the direction of rotation of the sub-frame, and the rotation angle, <span class="math-container">$\theta_A$</span>, is measured from the displaced member to its original position, which was set as the datum.</p>
<p><a href="https://i.stack.imgur.com/TCasX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TCasX.png" alt="enter image description here" /></a></p>
<p><strong>ADD:</strong> Hope this sketch helps to understand the equation and clear your question.</p>
<p><a href="https://i.stack.imgur.com/2jgVv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2jgVv.png" alt="enter image description here" /></a></p>
| 50143 | "$\theta_A$" and "$J_A \theta_A$" in the opposite direction |
2022-03-11T18:46:17.080 | <p><a href="https://i.stack.imgur.com/bKfhP.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bKfhP.jpg" alt="enter image description here" /></a></p>
<p>Especially on lower pitched roofs, this web member is practicality horizontal. Sure seems like it's providing minimal lateral support and runs the risk of falling. Is this a poor design, and if so would adding a vertical member underneath it make it better?</p>
| |structural-engineering| | <p>The added vertical member will improve the stability of the end panels as it decreases the L/r ratio of the bottom chord that it is attached to, however, it does not affect the distribution of internal forces since it is a "zero force member".</p>
| 50154 | Is this web member in the double fan truss design problematic? |
2022-03-13T20:56:48.833 | <p>when we plot the shear or moment diagram we add hachures to the area created with these diagrams , in most cases we see vertical hachures, is it represent the internal forces? or its just used to highlight the area ? and can we use inclined hachures?<a href="https://i.stack.imgur.com/EbzFn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EbzFn.jpg" alt="enter image description here" /></a></p>
| |statics|solid-mechanics|shear|bending|diagram| | <p>In moment and shear/axial force diagrams, the hachure is completely arbitrary, it can be however you want it to be. In fact, I'd say it's most common (outside of textbooks) to use no hachure at all, just the beam and the curve describing the internal force.</p>
<p><a href="https://i.stack.imgur.com/brsVO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/brsVO.png" alt="enter image description here" /></a></p>
<p>Most times what we see aren't hachures but step-diagrams, where we see the value of the diagram at a given interval:</p>
<p><a href="https://i.stack.imgur.com/kq2T8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kq2T8.png" alt="enter image description here" /></a></p>
<p>However, as <a href="https://engineering.stackexchange.com/questions/50167/should-the-hachures-in-moment-and-shear-diagram-be-vertical#comment93452_50167">@r13 correctly stated in their comment</a>, the opposite is true when drawing a distributed load, where the direction of the force is very relevant, so drawing the hachure properly is necessary:</p>
<p><a href="https://i.stack.imgur.com/Whj0F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Whj0F.png" alt="enter image description here" /></a></p>
| 50167 | should the hachures in moment and shear diagram be vertical? |
2022-03-13T22:09:53.637 | <p>It seems to be that the consensus behind impact drivers is that when the tool stalls, the hammer camms off the anvil, picks up speed turning 180 degrees, and then the moving hammers strike the anvil, losing all their energy, and using inertia of the hammers stopping to generate a pulse, a spike of torque.</p>
<p>However looking at some people operating example models or slowly operating an open driver module, it looks like as soon as the hammers cam over the anvil, and the spring picks up tension, in this moment, before the hammers do any lateral traveling, it looks like with no further rotational input, the spring releases, and the hammers surge in speed, with no assistance from the motor.</p>
<p>You can observe this when someone is operating a driver without a motor, just with manual operation. As soon as the spring is released, and they arent moving their hands, the hammer jolt forward.</p>
<p>So it looks like the spring actually stores energy for impact purposes, and it transfers it's compression into rotational force.</p>
<p>My question is how? It's not very clear and no one shines any light on it or brings it to attention or shows the inside of the impact driver's walls.</p>
<p>Everyone just explains that the motor drivers the hammers and ignores the hammers launching themselves via the spring.</p>
<p>Does the inside of the impact module have groves, is there another piece of metal the spring rests on that is inside those groves?</p>
<p>So is it the case that when the motor spins, it drives the hammers, when the driver stalls, the motor drivers the spring instead, the spring becoming a capacitor for kinetic energy whenever the output shaft isn't moving? So half the time the hammers get power, half the time the spring gets power? Then the output shaft benefits from the motor and the springs combining momentum?</p>
<p>How would you double the torque? put in a stiffer spring?
Couldn't you put in a 2000 pound spring and use 20 times torque reduction on the motor to make a metal shattering tool? that would probably destroy itself.</p>
| |impact| | <p><a href="https://i.stack.imgur.com/Jjzqh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jjzqh.png" alt="enter image description here" /></a>
<a href="https://www.wonkeedonkeetools.co.uk/cordless-impact-drivers/how-does-the-impact-function-work" rel="nofollow noreferrer">https://www.wonkeedonkeetools.co.uk/cordless-impact-drivers/how-does-the-impact-function-work</a></p>
<p>Coil springs can be both compressed and twisted.</p>
<p>The compression of the spring is what allows the hammer to cam out. The twisting or winding up of the spring is what stores the energy for the rotational impact.</p>
<p>That's why the ramp that the hammer cams out on is so steep. If you were just wanting the hammer to cam out so that the motor spin unloaded to pick up speed to slam into the anvil, you could use a gentler ramp. But if you're trying to wind up the spring you need a much steeper ramp so the motor can continue to spin while the hammer stays still as much as possible without actually remaining still so that it can eventually unwind.</p>
<blockquote>
<p>How would you double the torque? put in a stiffer spring? Couldn't you put in a 2000 pound spring and use 20 times torque reduction on the motor to make a metal shattering tool? that would probably destroy itself.</p>
</blockquote>
<p>No. For all intents and purposes, it would just behave like a a regular drill which has no camming out all because your wrists wouldn't be able to stop the drill from rotating when it is supposed to be camming out. It would just torque and break your wrists like a high torque drill would. It wouldn't even shatter your wrists because no camming would occur and thus no shattering impact.</p>
| 50168 | How does an impact driver convert the lateral motion of the spring expansion, into rotational force? |
2022-03-14T10:52:00.957 | <p>I'm in a use case where the outer surface is a gear driven by a motor (that's the easy part) and the inner surface meshes with a metallic (iron) rope. When the part rotates, I want the motor and everything to climb the rope <strong>the same way a nut moves in a bolt</strong>. The part and the rope share the same cylindrical axis.</p>
<p>I want the part to be as simple as possible, no assembly required. I also need the part to be as general as possible. Let's say the pitch of the rope is <strong>p</strong> mm and the angle is <strong>theta</strong> degrees.</p>
<p><img src="https://image.shutterstock.com/image-photo/collection-various-ropes-string-on-260nw-757945858.jpg" alt="Rope" /></p>
<p>If there's any more details required, I'm happy to elaborate. I've brainstormed for a few hours and I'm coming up blank. Best I can do is doing a 3D negative of the rope surface. In theory it will climb but in practice, it will be unreliable and easily wear out</p>
| |design|mechanical|geometry| | <p>I believe this answer is mostly applicable to your question, although the "assembly" would be ninety degrees rotated. I've recently learned of a system known as <a href="https://www.sdp-si.com/products/Timing-Belts-and-Cables/Synchromesh-Drive-Systems.php" rel="nofollow noreferrer">Synchromesh Drive Systems</a> which involves a straight wire wrapped with a helical wire, making up the drive cable, and a matching pulley.</p>
<p><a href="https://i.stack.imgur.com/W57dw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/W57dw.png" alt="synchromesh drive pulleys" /></a></p>
<p><a href="https://i.stack.imgur.com/zUeSV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zUeSV.png" alt="synchromesh drive cable." /></a></p>
<p>Photos from linked site. There is also a video showing this type of drive system being used in a 3D printer.</p>
<p>One could use this system for climbing if the cable was fixed securely and the motor was part of a traveler, designed to maintain cable contact, perhaps with tensioning pulleys on the inbound and outbound sides of the drive pulley.</p>
<p>If one considers the typical wire rope, the alignment amounts to a very steep pitch, making the mechanical aspect quite challenging. A wire rope with a smaller pitch akin to a screw is going to be more like a spring than a rope. Combining a conventional wire rope with a "spring wound rope" begins to approximate the Synchromesh Drive System.</p>
| 50176 | Is there a geometry that meshes exactly with a rope? |
2022-03-14T11:18:49.887 | <p>Consider a simply supported beam, acted upon by a concentrated load P as shown</p>
<p><a href="https://i.stack.imgur.com/q48ol.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q48ol.jpg" alt="enter image description here" /></a></p>
<p>By applying the method of successive integration, i.e. by using the differential equation</p>
<p><span class="math-container">$$EI \frac{d^2v}{dx^2} = M$$</span></p>
<p>We can findd out the equation (i.e. <span class="math-container">$v= f(x)$</span> ) for the deflection curve:</p>
<p><a href="https://i.stack.imgur.com/wKvY1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wKvY1.jpg" alt="enter image description here" /></a></p>
<p>I was interested in finding out the maximum deflection of the beam, which can be done by setting</p>
<p><span class="math-container">$$\frac{dv}{dx}=0$$</span></p>
<p>However, there are two equations for v, one for interval 0 to a and second for a to L, which equation should I differentiate to obtain the maximum deflection?</p>
<p>The book that I follow, differentiates the first equation for v (0 to a), I don't understand why though.</p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam| | <p>As stated in the question, the equation for the deflection curves for the intervals <span class="math-container">$0<x<a$</span> and <span class="math-container">$a<x<L$</span> are given as</p>
<p><a href="https://i.stack.imgur.com/el0Z9.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/el0Z9.jpg" alt="enter image description here" /></a></p>
<p>The maximum deflection in the beam can occur either to the left of the load or to the right, and we don't know what is the case here.</p>
<p>Let us start off by assuming the maximum deflection was in the portion <span class="math-container">$0<x<a$</span>. If that were to be true then at some x lying between 0 to a the slope of the deflection curve for <span class="math-container">$0<x<a$</span> would've been zero. That x at which slope would've been zero can be determined by differentiating the equation for deflection curve for <span class="math-container">$0<x<a$</span> and then setting it to 0. This would give us,</p>
<p><span class="math-container">$$x= \sqrt{\frac{L^2 - b^2}{3}}$$</span></p>
<p>If this indeed is the value of x, at which the deflection is maximum and if it occurs in <span class="math-container">$0<x<a$</span>, then <span class="math-container">$x<a$</span>. If x turns out to be greater than a, that would mean this result is not feasibe. Let us try to investigate whether this x is greater than or less than a.</p>
<p><span class="math-container">$$x= \sqrt{\frac{L^2 - b^2}{3}}$$</span></p>
<p><span class="math-container">$$x= \sqrt{\frac{a^2+2ab}{3}}$$</span></p>
<p><span class="math-container">$$\sqrt\frac{a^2+2a.a}{3} > \sqrt\frac{a^2+2ab}{3}$$</span></p>
<p><span class="math-container">$$a > \sqrt\frac{a^2+2ab}{3}$$</span></p>
<p><span class="math-container">$$a > x$$</span></p>
<p>The value of x, indeed is in the interval 0<x<a. Thus, our assumption was right and the maximum deflection does occur in the region 0<x<a. To find this deflection we can substitute <span class="math-container">$$x= \sqrt{\frac{L^2 - b^2}{3}}$$</span></p>
<p>in the equation of the deflection curve for <span class="math-container">$0<x<a$</span></p>
<p>We could extend the analysis and disprove that the maximum deflection occurs in the interval <span class="math-container">$a<x<L$</span>. In the same manner, we can first assume that the max deflection was occurring in the portion <span class="math-container">$a<x<L$</span>, then we can determine the x at which it would've occurred (if it would've occurred), by differentiating and putting to 0, the equation of the deflection curve for <span class="math-container">$a<x<L$</span>.</p>
<p>The equation thus obtained will be a qaudratic one, which has solutions</p>
<p><a href="https://i.stack.imgur.com/eT1hf.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eT1hf.jpg" alt="enter image description here" /></a></p>
<p>The second value of x is greater than L, hence this is not a feasible solution. Also, by using inequalities we can prove that the first value of x, will be <a. Since this value is <a, we conclude that even the first value of x in the solution above is not feasible since x should occur between <span class="math-container">$a<x<L$</span> if the max deflection were to truly occur in the region <span class="math-container">$a<x<L$</span>.</p>
<p>Here is a plot of the deflection curves for a 6m length of the beam-</p>
<p><a href="https://i.stack.imgur.com/OfjhH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OfjhH.jpg" alt="enter image description here" /></a></p>
| 50177 | Maximum deflection in the beam |
2022-03-15T17:51:27.703 | <p>I would like to know if a very bright light will attract an incoming laser guided weapon such as a laser guided missile.</p>
<p>Say for example that a tank has a six-foot pole on the top of it and attached to the end of this pole is a high-voltage carbon arc lamp and it emits light in all directions. I am wondering if the intense light of a carbon arc lamp will outshine the reflecting laser light coming off the surface of the tank and the incoming laser guided missile will steer itself towards the carbon arc lamp.</p>
<p><a href="https://i.stack.imgur.com/YVwTy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YVwTy.jpg" alt="enter image description here" /></a></p>
<p>Can a very bright light attract an incoming laser guided weapon?</p>
<p>I want to point out that I am not an engineer and I know very little about how laser guided systems work, so I am asking this question simply out of scientific curiosity.</p>
| |mechanical-engineering|electrical-engineering|optics|lasers| | <h2>Probably not.</h2>
<p>Look, anything is possible. You could throw some rocks at a keyboard, and have it open SE, log in, and post an proof of trisection.</p>
<p>It ain't real likely.</p>
<p>Laser guided weapons usually have some specific frequency they're looking for. Then, on top of that, the laser is pulsed as an identifier (think of it spelling four letters in Morse code on a loop — not how it works, but a good analogy).</p>
<p>To misguide your missile, you're looking at the narrow band of light you're looking for coming off the arc lamp, then flickering in a way that duplicates the "Morse code" the weapon is looking for. That's... unlikely.</p>
<p>More plausible is that you can put out so much light that you blind the missile, and it doesn't hit anything. Of course, the more traditional way to do this is with another laser that you shine back at it - much lower power cost; far fewer friendly effects.</p>
| 50192 | Can a very bright light attract an incoming laser guided weapon? |
2022-03-16T12:56:19.283 | <p>I am performing extraction flash calculations for 4 component and 2 phase system in Matlab. For anyone somewhat shaky with what extraction flash calculation is; extraction is performed, feed composition is known and we are calculating compositions of both phases at equilibrium, mole fraction of every component in both phases as well as mole fraction of both phases.</p>
<p>This calculation is done numerically, until convergence criterion is satisfied. I know it is possible to take calculated mole fraction difference in two consecutive iterations as a convergence criterion, but I am not sure how many criteria I should apply in case of 4 component and 2 phase system?</p>
<p>I am thinking I need to apply 3 such criteria for 3 components in the system for one of the phases because mole fraction of 4th component is satisfied automatically by mass balance for the phase and mole fractions of components in other phase is satisfied automatically by phase equilibrium condition or mass balance for every component.</p>
<p>I am not sure if this is correct?</p>
| |chemical-engineering|matlab| | <p>For reference, I am drawing from the chapter on isothermal VLE separations in Separation Process Principles by Seader, Henley, and Roper.</p>
<p>Consider only the final state. The two starting statements in degrees of freedom with <span class="math-container">$C'$</span> independent chemical components and <span class="math-container">$\Pi_{act}$</span> phases give the number of intensive parameters and independent intensive equations.</p>
<p><span class="math-container">$$\nu = C'\Pi_{act} + 2$$</span></p>
<p><span class="math-container">$$\epsilon = \Pi_{act} + C'(\Pi_{act} - 1)$$</span></p>
<p>For your system with <span class="math-container">$C' = 4$</span> and <span class="math-container">$\Pi_{act} = 2$</span>, we find <span class="math-container">$\nu = 10$</span> and <span class="math-container">$\epsilon = 6$</span>. The parameters are</p>
<p><span class="math-container">$$T, p, x_j(4), y_j(4)$$</span></p>
<p>The equations are</p>
<p><span class="math-container">$$\sum x_j = 1, \sum y_j = 1, K_j(T, p)(4)$$</span></p>
<p>Assuming that you have theoretical equations for partition functions <span class="math-container">$K_j(T, p) = y_j/x_i$</span>, you will only have to specify four parameters at the end state. These generally are taken as <span class="math-container">$T, p$</span> (allowing you to obtain <span class="math-container">$K_j$</span>) and two composition variables. When you do not have pre-defined theoretical equations for the partition coefficients <span class="math-container">$K_j(T,p)$</span>, you drop three independent equations (because all four <span class="math-container">$K_j$</span> are not independent). You will have to measure three additional compositions to allow you to construct the <span class="math-container">$K_j$</span> values empirically.</p>
<p>When we do an isothermal flash from <span class="math-container">$N_F$</span> feed streams, we have these additions to the number of parameters and equations</p>
<p><span class="math-container">$$\Delta \nu = N_F(C' + 1) + \Pi_act$$</span>
<span class="math-container">$$\Delta \epsilon = N_F + C' + 1 $$</span></p>
<p>In your case, the flash adds initial compositions <span class="math-container">$z_j(4)$</span>, initial phase amount <span class="math-container">$f_{z}$</span>, and ending phase fractions <span class="math-container">$f_L, f_V$</span> as intensive parameters. You add these six independent equations</p>
<p><span class="math-container">$$\sum z_j = 1, f_z = 1, f_L + f_V = 1, z_j f_z = x_j f_L + y_j f_V$$</span></p>
<p>The end result is that you must measure one further intensive parameter from the flash, typically the vapor fraction <span class="math-container">$f_V$</span>.</p>
<p>Finally, a two phase flash is directly amenable to being set up for an analytical solution using a Rachford-Rice expression.</p>
<p><span class="math-container">$$\sum \frac{z_j\left(1 - K_j\right)}{1 - f_V\left(1 - K_j\right)} = 0$$</span></p>
<p>Graphical analysis is easier than an iterative analysis.</p>
<p>The four problems that tend to crash out when doing (isothermal) flash calculations are as follows:</p>
<ul>
<li><p>Under specifying measured parameters -- measuring too few parameters to comply with the requirement from the difference between known independent equations and known intensive parameters <span class="math-container">$(\epsilon + \Delta \epsilon) - (\nu + \Delta \nu)$</span>.</p>
</li>
<li><p>Incorrectly specifying independent equations -- not using the correct equations or adding additional parameters in the equations. Specific examples that generally fail here involve incorrect or additional complex equations for partition functions.</p>
</li>
<li><p>Improperly specifying parameters -- counting measured parameters that are not truly independent as part of the "independent" set. Specific examples that fail here involve measuring mole fractions for all components in one stream but ignoring any measurements in another stream.</p>
</li>
<li><p>Over specifying parameters -- trying to analyze the system after pre-specifying more parameters than permitted by the degrees of freedom.</p>
</li>
</ul>
<p>You might be able to judge from this which failure you happen to be approaching in your analysis.</p>
| 50196 | Extraction Flash Calculation - 4 Component and 2 Phase System |
2022-03-16T21:45:15.517 | <p>In a race car rear wing with 2 airfoils, why does the second airfoil have a greater angle of attack compared to the leading airfoil? Also, what is the best offset distance between the 2 airfoils in a racing car rear wing?</p>
| |airflow| | <p>Airfoil creates lift (or in this case downward force) by bending the airflow downward, (upward here).</p>
<p>The lift coefficient and hence lift force is almost linearly proportional to the angle of attack up to approximately 10 degrees.</p>
<p>So the second airfoil needs to pitch a bit more to keep the same angle of attack with a relative wind that is already bent by the first airfoil.</p>
<p>The offset distance is determined in a wind tunnel.</p>
<p><a href="https://i.stack.imgur.com/DiR2g.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DiR2g.png" alt="lift vs AOA" /></a></p>
| 50203 | Airflow theory in wing design |
2022-03-18T17:52:05.957 | <p>Is it reasonable to operate a fully mechanical heat pump? For example, using a linear back-and-forth motion from a vortex bladeless turbine to operate a mechanical compressor to heat (or cool off) a sizeable container of water?</p>
| |heat-transfer|renewable-energy|energy-storage| | <p>Virtually every possible way of doing this is already out there.</p>
<p>For bolt-ons, you can find propshaft pto chain drives and belt drives, hydraulic and mechanical engine pto systems, and front of block bolt-ons that are fan belt driven or bolted to the propshaft head.</p>
<p>Stand alone systems are available from suitecase sized to boxcar sized.</p>
<p><a href="https://www.yanmar.com/global/energy/ghp/" rel="nofollow noreferrer">https://www.yanmar.com/global/energy/ghp/</a></p>
<p>We expect that you do some research before you ask here. Any google search will give you many examples. There are millions of these out there used where shaft power is readily available (like on ships) or where electricity is scarce or sketchy.</p>
<p>Some references for further research.</p>
<blockquote>
<p>Sanaye et al. (2010) analysed a GHP for the Iranian climate conditions and found that depending on climate and demand, annual operating costs (which is also dependent on energy price) for EHPs were higher than that GHPs. They also found that GHPs were more economical than EHPs in the residential sector. Even GHP air conditioning systems compared to the same kind of EHP air conditioning system were more economical (Lian et al., 2005) . Engine driven heat pump systems have been used for different industrial applications (Gungor et al., 2011) but few studies have been done on domestic applications of an engine driven heat pump. Hepbasli et al. (2009) performed a review and mathematical modelling for GHP systems for residential and industrial applications. Various applications and different arrangements of GHP have been discussed. They found GHP systems are more efficient when used for both water and space heating.</p>
</blockquote>
<p><a href="https://www.researchgate.net/publication/269393425_A_REVIEW_OF_DIESEL_ENGINE_DRIVEN_HEAT_PUMPS_DEHPs_AND_THEIR_POTENTIAL_IN_THE_UK_DOMESTIC_SECTOR" rel="nofollow noreferrer">https://www.researchgate.net/publication/269393425_A_REVIEW_OF_DIESEL_ENGINE_DRIVEN_HEAT_PUMPS_DEHPs_AND_THEIR_POTENTIAL_IN_THE_UK_DOMESTIC_SECTOR</a></p>
| 50227 | Does a fully mechanical heat pump make sense? |
2022-03-19T11:19:02.257 | <p>This is a pivot point on a linkage in my coffee table. It appears to be a rivet with a tail that was pressed with a cross-shaped punch to widen it. I haven't see something like this before. Is there a name for this? What is the benefit of cross-punching it?<a href="https://i.stack.imgur.com/2NiEU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2NiEU.jpg" alt="enter image description here" /></a></p>
| |fasteners|rivets| | <p>It was hit with a cross-shaped punch to get that side or head to expand.</p>
<p>Usually done with a single blow for speed and low cost. Other choices can be nuts and bolts with nylock nuts so they don’t come loose easily.</p>
| 50239 | What type of rivet is this? |
2022-03-19T19:58:20.850 | <p>I am building this ground-mounted solar panel sail:</p>
<p><a href="https://i.stack.imgur.com/kdjaS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kdjaS.jpg" alt="enter image description here" /></a></p>
<p>It will sit on twelve 150 mm timber piles. There are 20 panels 1755×1038 mm, 20 kg each, with 17 mm gaps between them, sitting at 69°. The total weight above the timber (panels + aluminium frame) is roughly 550 kg.</p>
<p>The piles are 2.4 m long, and they will be in the ground 1.2–1.5 m deep.</p>
<p>The triangle base is 2.23 m, height 2.91 m and side 3.11 m.</p>
<p>The question is <strong>how heavy the piles' feet</strong> (not on the picture) need to be to stop the sail from flying away when it storms. The goal is to ensure that a storm would rather rip the structure apart than lift/tip/fly it. Occasionally winds may top 150 km/h in this area.</p>
<p>I've asked a professional structural engineer this question and he's given me this figure of at least 0.6 m<sup>3</sup> of concrete per pile.</p>
<p>Assuming density 2.4 t/m<sup>3</sup>, that means a weight of more than 17 tonnes is required to hold the structure down.</p>
<p>Does the weight figure not sound overestimated? Would a wind be able to come any close to lifting that weight by blowing in the back of this sail before it rips it apart?</p>
<p>Also, will the cross beams between the piles do a good job? My assumption was that they should provide robustness to the structure so that it should not be possible to lift the rear row of piles without lifting the front one to the same distance.</p>
| |structural-engineering|civil-engineering|wind|pile-foundations| | <p>Another thing you need to consider, is the <strong>maximum wind pressure</strong> that the panels will experience for the calculation -- much depends on this. In the other answers, it's not explicitly mentioned what value should be used.</p>
<p>Solar panels are built to standard specifications. Usually, most solar panels are built to sustain a 200 -300 kg/m^2 Wind load. The actual wind loading value is bound to be in the datasheet. That value should be used (instead of calculated value from nominal airpressure <span class="math-container">$q_n=\frac{1}{2}\rho v^2$</span>, which is then increased by safety factors).</p>
<p><strong>The reason is that if the calculated air pressure exceeds the specified wind loading then the panel will break off the structure.</strong></p>
<p>Even in the unlikely event that the specified wind loading is less than the total calculated air pressure (this is doubtful because the quoted wind speed of 150 km/hr is quite high -- it corresponds to 41 m/s when Eurocode annexes suggest in the absence of data a maximum average velocity of 25 or in very windy areas 33 m/s), using the maximum value that the panels can sustain is a safer approach.</p>
| 50246 | How strong will this sail pull up? |
2022-03-21T18:03:21.753 | <p>When a beam which is fixed at one end and free at the other, is acted upon by a load P, the beam bends and we get a bending moment at every cross section of the beam. This bending moment can be determined by making an imaginary cut at the cross section where BM is to determined, and then applying a moment balance on either parts of the beam obtained after the cut.</p>
<p><a href="https://i.stack.imgur.com/JeYGW.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JeYGW.jpg" alt="enter image description here" /></a></p>
<p>Now consider a rigid bar fixed from one of its ends and a load P is applied on the other. Will there be any bending moment in this case? I mean, if we cut the bar from a section we can still perform a moment balance and that would require that some moment be developed at the section to bring the parts of the being in equilibrium. So does that mean we would still get a bending moment even if the bar is rigid?</p>
| |mechanical-engineering|structural-engineering|structural-analysis|beam| | <p>Yes, a rigid beam can contain bending moments (as well as any other internal forces). Stiffness affects how an element deforms, not its capacity to resist forces.</p>
<p>In isostatic (statically determinate) structures, the stiffness is in fact entirely irrelevant when determining internal forces.</p>
<p>In hyperstatic (statically indeterminate) structures, the behavior depends on whether all elements have the same stiffness. If they do, then the stiffness is also irrelevant and "cancels out".* However, if beams have different stiffnesses (i.e. one is rigid and the other flexible), then the stiffer beam will "pull" more internal forces.</p>
<hr />
<p>* <sub>Mathematically, a statically indeterminate structure where all elements are perfectly stiff actually can't be solved. That's because hyperstatic structures are solved via compatibility equations for deflection and rotation. But if everything's perfectly stiff, the equations can give any solution you want, since the deflection will always be zero.<br />
However, if we remember that this is just a model, and in reality there's no such thing as a perfectly stiff beam, then we can forgive the mathematical sin of cancelling out the infinite stiffness and be on our merry way.</sub></p>
| 50274 | Rigidity and bending moment |
2022-03-22T12:46:11.067 | <p>i am trying to find out the best way to approach a design task for a plane wing with a truss support. i am treating the wing as a simply supported beam but i have no idea how to incorporate the truss except as and extra support but i don't know whether this is suitable.</p>
<p>i am assuming it can be done via truss and member methods, this is the layout i am attempting</p>
<p><a href="https://i.stack.imgur.com/Nw6F7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Nw6F7.png" alt="enter image description here" /></a></p>
| |beam|aerospace-engineering| | <p>Your wing can be assumed roughly as a two-span continuous beam supported from left by pin support (or fixed support but more on that later), then by other pin support by the diagonal tension strut, and the last right part as a cantilever beam.</p>
<p>Assuming the distributed load of W upward
WE have the maximum moment above the strut at the middle of the beam</p>
<p><span class="math-container">$$M_{L2}=-WL_2^2/$$</span></p>
<p><span class="math-container">$$M_{L1}= -WL_1^2/8+1/2M_{L2}$$</span>
And the tensin in strut is</p>
<p><span class="math-container">$$T=sec( a )*(1/2 \ W*L1+W*L2)$$</span></p>
<p>'</p>
<p><a href="https://i.stack.imgur.com/z9xE8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z9xE8.jpg" alt="wing" /></a></p>
| 50290 | Simply supported beam with a truss |
2022-03-22T21:50:37.330 | <p>In an engineering drawing how would you dimension the cross-section of an aircraft's airfoil? Obviously, you would have to include the chord length, leading and trailing edge radius, and wingspan. But this does not fully define the geometry.</p>
| |technical-drawing| | <p>A search using "airfoil design parameters" returned at the top of the list a <a href="https://drahmednagib.com/onewebmedia/SPC307-2017/SPC307-Lec.8.pdf" rel="nofollow noreferrer">link to a PDF</a> with specifics regarding this question. It appears to be a 72 slide Powerpoint presentation, but one page narrows the information:</p>
<p>Some of the basic parameters to describe the airfoil geometry are:</p>
<ol>
<li>Leading edge—the forward most point on the airfoil (typically placed at the origin
for convenience)</li>
<li>Trailing edge—the aft most point on the airfoil (typically placed on the x axis for
convenience)</li>
<li>Chord line—a straight line between the leading and trailing edges (the x axis for our
convention)</li>
<li>Mean camber line—a line midway between the upper and lower surfaces at each
chord-wise position</li>
<li>Maximum camber—the largest value of the distance between the mean camber
line and the chord line, which quantifies the camber of an airfoil</li>
<li>Maximum thickness—the largest value of the distance between the upper and
lower surfaces, which quantifies the thickness of the airfoil</li>
<li>Leading-edge radius—the radius of a circle that fits the leading-edge curvature</li>
</ol>
<p>The document contains images to reference the terms above as well as detailed explanations for these parameters.</p>
| 50295 | Airfoil engineering drawing |
2022-03-23T02:07:02.710 | <p>I'm working on a thermal expansion problem out of Beer & Johnston's <em><a href="https://www.mheducation.com/highered/product/mechanics-materials-beer-johnston/M9781260113273.html" rel="nofollow noreferrer">Mechanics of Materials</a></em> and I'm a bit confused by the underlying rationale of one of the steps in the solution process.</p>
<p><a href="https://i.stack.imgur.com/GOTKA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GOTKA.png" alt="enter image description here" /></a></p>
<p>In the exercise text, <span class="math-container">$\alpha_s$</span> and <span class="math-container">$\alpha_b$</span> are given along with a <span class="math-container">$\sigma_{max}$</span> of the inner steel core that cannot be exceeded. The core is considered fully bonded to the shell and the goal is to derive the max <span class="math-container">$\Delta T$</span> that can be observed while remaining under <span class="math-container">$\sigma_{max}$</span>. I understand the general solution process, but a sign change in the deformation equation has me confused, as the book does not give a specific rationale for its change. The equation the text gives is:</p>
<p><span class="math-container">$$\delta = \alpha (\Delta T) L + \frac{PL}{EA}$$</span></p>
<p>I have verified the final solution with a solutions manual, and the solution process for this problem proceeds:</p>
<p><span class="math-container">$$\begin{align}
P &= A_s \sigma_{max} & (1) \\
\delta_b &= \alpha_b (\Delta T) L - \frac{PL}{E_b A_b} & (2) \\
\delta_s &= \alpha_s (\Delta T) L + \frac{PL}{E_s A_s} & (3) \\
& \text{solve for} \,\,\Delta T \,\, \text{below}
\end{align}$$</span></p>
<p>The specific element in question is the sign change of the <span class="math-container">$\delta_P$</span> term in Equation (2). In particular,</p>
<ol>
<li>Why does this sign change occur?</li>
<li>For what reason is it assigned to the brass shell and not the inner steel core?</li>
</ol>
| |solid-mechanics|thermal-expansion|homework| | <p>The solution depends on the coefficients of thermal expansion (CTE) for brass (<span class="math-container">$\alpha_b$</span>) and steel (<span class="math-container">$\alpha_s$</span>) and whether the temperature is increasing (<span class="math-container">$\Delta T>0$</span>) or decreasing (<span class="math-container">$\Delta T<0$</span>).</p>
<ul>
<li>If <span class="math-container">$\Delta T$</span> is positive, both will expand due to the temperature. This is the term <span class="math-container">$+\alpha (\Delta T) L$</span> in both equations.</li>
<li>If brass has a larger CTE than steel, the larger expansion of the brass will put the steel in tension. The resulting force P will cause the steel to get longer than the thermal expansion alone. This is the term <span class="math-container">$+ \frac{PL}{E_s A_s}$</span>.</li>
<li>At the same time, the steel will put the brass in compression. The resulting force P will cause the brass to get shorter than due to thermal expansion alone. This is the term <span class="math-container">$- \frac{PL}{E_b A_b}$</span>.</li>
</ul>
<p>In summary, the sign change is because one part is in compression due to the force P and the other part is in tension.</p>
| 50299 | Derivation of deformation equation given thermal expansion |
2022-03-23T05:11:29.653 | <p><a href="https://i.stack.imgur.com/hxb0tl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hxb0tl.jpg" alt="![enter image description here" /></a></p>
<p>Here, <span class="math-container">$P>Q$</span>. <span class="math-container">$O$</span> is the center of mass of the rigid and uniform bar/stick.</p>
<p>As <span class="math-container">$P>Q$</span>, the resultant is situated to the right of <span class="math-container">$\vec{P}$</span> and is parallel to <span class="math-container">$\vec{P}$</span>. The magnitude of the resultant is <span class="math-container">$P-Q$</span>.</p>
<p>To convince you that the figure is correct, I'll do some math to prove it.</p>
<p>Let us obtain the sum of torques about the center of mass,</p>
<p><a href="https://i.stack.imgur.com/kxCfFl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kxCfFl.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$(P-Q)b=Pa+Qa$$</span></p>
<p><span class="math-container">$$b=\frac{P+Q}{P-Q}a$$</span></p>
<p><span class="math-container">$$b=fa\ \left[\text{Let $f=\frac{P+Q}{P-Q}$}\right]$$</span></p>
<p>As <span class="math-container">$P>Q$</span>, <span class="math-container">$f>1$</span>, and <span class="math-container">$b>a$</span>. So, the correct figure will be,</p>
<p><a href="https://i.stack.imgur.com/sMtVHl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sMtVHl.jpg" alt="enter image description here" /></a></p>
<p>I hope you're satisfied that the figure is correct.</p>
<p><strong>My comments:</strong></p>
<p>Is it possible to replace <span class="math-container">$\vec{P}$</span> and <span class="math-container">$\vec{Q}$</span> with a single force? I mean practically, not theoretically. From the figure, we can see that the resultant force is outside the bar. In other words, <span class="math-container">$\vec{P}$</span> and <span class="math-container">$\vec{Q}$</span> can be replaced by a force of magnitude <span class="math-container">$P-Q$</span>, which will act outside the bar. This may be possible theoretically; however, this is not possible practically as the resultant force will be acting on literally nothing as it is outside the bar. Therefore, I conclude that it is impossible to replace <span class="math-container">$\vec{P}$</span> and <span class="math-container">$\vec{Q}$</span> with a single force practically. Theoretically, it is possible, but practically, no.</p>
<p><strong>My question:</strong></p>
<ol>
<li>Can <span class="math-container">$\vec{P}$</span> and <span class="math-container">$\vec{Q}$</span> be replaced by a single force? Is my conclusion correct?</li>
</ol>
<hr />
<p>These may help you to answer this question:</p>
<ol>
<li><a href="https://math.stackexchange.com/questions/4409056/how-will-the-force-act-outside-the-stick?noredirect=1#comment9224729_4409056">Comment by @Ivan</a></li>
<li><a href="https://physics.stackexchange.com/a/699905/260477">Answer by @Farcher</a></li>
</ol>
<hr />
<p>This question was posted with the help of <a href="https://physics.stackexchange.com/users/196140/eli">@Eli</a>.</p>
| |mechanical-engineering|applied-mechanics|torque|statics|dynamics| | <p>As you have calculated the torque on the bar is</p>
<p><span class="math-container">$$\tau= (P+q)A$$</span></p>
<p>and a net force</p>
<p><span class="math-container">$$F=P-Q$$</span></p>
<p>This will cause the bar to turn with an angular acceleration,</p>
<p><span class="math-container">$$\alpha=\frac{\tau}{I}$$</span></p>
<p>and also accelerate with,</p>
<p><span class="math-container">$$a=\frac{P-Q}{m}$$</span></p>
<p>Any substitute pair of forces acting within the length of the bar can be scaled by the factor of <span class="math-container">$A/D$</span> to impart the same torque. But the new net force will not be the same.</p>
<p><span class="math-container">$P_N-Q_N\neq P-Q$</span>.</p>
<ul>
<li>A= half-length of bar</li>
<li>m= mass</li>
<li>a= linear acceleration</li>
<li>D = distance of new pair of force Pn, Qn, from the center of the bar</li>
<li><span class="math-container">$\alpha$</span>= angular acceleration</li>
<li>I= bar's moment of inertia</li>
<li><span class="math-container">$\tau$</span>= torque</li>
</ul>
<p>So depending on what you demand the answer varies, if you require just the same torque, yes. If you require the same torque and linear acceleration no!</p>
| 50300 | How to replace these two forces with one force? |
2022-03-24T05:53:12.300 | <p>Say the cross section carries a vertical shearing force of 40 kN. And say I want to find the value of the shear stress just below the section a-a.</p>
<p>I am very confused about which area <span class="math-container">$(A)$</span> to take When calculating <span class="math-container">$Q = y \times A$</span> in the <span class="math-container">$\tau = \frac{VQ}{It}$</span>, shear force equation. I have drawn 2 sketches and can someone please explain which one is correct?</p>
<p>Also where is the maximum shear in this case, I assume in a location where the width is small? ie the upper rectangle?</p>
<p><a href="https://i.stack.imgur.com/VJ9DT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VJ9DT.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/kPXtD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kPXtD.png" alt="enter image description here" /></a></p>
| |structural-engineering|structural-analysis|shear| | <p>By definition shear flow on a beam's cross-section is the same on the two sides of any horizontal plane.</p>
<p>eg, the top surface is trying to move to the right, the bottom surface is reacting by trying to move to the left with an equal amount of shear to maintain equilibrium. Otherwise, the beam will start to move to the left or right.</p>
| 50311 | How to find the shear stress just below or just above some elevation? |
2022-03-25T08:54:21.683 | <p>I'm reading about static determinacy of structures <a href="https://learnaboutstructures.com/External-Indeterminacy" rel="nofollow noreferrer">here</a>.</p>
<p>A quote:</p>
<blockquote>
<p>Therefore, one extra equilibrium equation is added to the system due to the introduction of the hinge: either ∑MABB=0 or ∑MBCB=0. Only one of these equations counts because the two equations are not independent. They are not independent because they both mean the same thing, that the moment at the hinge is zero. If I know the moment on one side of the cut is zero, then I know automatically as well that the moment on the other side of the cut must be zero. So, for each internal hinge in a structure, there is a single equation of condition: ec=1.</p>
</blockquote>
<p>So, the introduction of a hinge gives us an extra equation, <span class="math-container">$\sum M_B=0$</span>. This is called an equation of condition. The text mentions that only a sum of moments on <em>one</em> side of the hinge counts as a new equation, as the moment sums on either sides mean the same thing.</p>
<p>Let's say we have a frame like this:</p>
<p><a href="https://i.stack.imgur.com/Rizxo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rizxo.png" alt="enter image description here" /></a></p>
<p>The frame has a hinge 1/3 of the way along the top beam. So if the reaction forces at the supports are <span class="math-container">$A_y$</span>, <span class="math-container">$A_x$</span>, and <span class="math-container">$B_y$</span> (roller support at B), we could write moment about the hinge in two ways:</p>
<p><span class="math-container">$$A_y \frac{L}{3} - A_x L = 0$$</span> or</p>
<p><span class="math-container">$$B_y \frac{2L}{3} = 0$$</span></p>
<p>Here I have neglected the loadings on the frame, but they would just be included as constants in the equations.</p>
<p>So what I don't get is why these equations cannot be used as <em>two</em> new equations. The text says they must be <strong>dependent</strong>. Why?</p>
<p>I'm not asking specifically about this example, but rather in general. If we derive an equation for the moment on both sides of the hinge, we do indeed calculate the same thing, but we have two <em>equations</em> for the same thing. So why don't they count as two equations for static determinacy?</p>
| |structural-engineering|statics| | <p>The sum of moments about any point is 0. Since 0=0 is always true, the equation for moments about different points turn out to be the same equation! This can be better understood from an example.
<a href="https://i.stack.imgur.com/7oYxI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7oYxI.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$\sum M_A=0 \Rightarrow (L_1F_1+L_2F_2 + ...) - aB = 0 \tag{1}$$</span>
<span class="math-container">$$\sum F=0 \Rightarrow B=A-(F_1+F_2+...) \tag{2}$$</span>
<span class="math-container">$$\sum M_B=0 \Rightarrow ((L_1+a)F_1 + (L_2+a)F_2 + ...) -aA = 0 \tag{3}$$</span>
Rearranging (3) and substituting (2) in the last step gives the identical equation as (1). Hence, (1) and (3) are the same equation, and both equations cannot be used to solve for the reaction forces.
<span class="math-container">$$(L_1F_1+L_2F_2+...) +a(F_1+F_2+...)-aA=0 \\
(L_1F_1+L_2F_2+...) +a(F_1+F_2+...-A)=0 \\
(L_1F_1+L_2F_2+...) -a(A-(F_1+F_2+...))=0 \\
(L_1F_1+L_2F_2+...) -aB =0
$$</span></p>
| 50326 | Equations of condition: why are two moment equations around a hinge dependent? |
2022-03-25T16:53:10.740 | <p>If flow separate in high AoA diffuser, vortex generators can help to delay separation.</p>
<p>Where is best palce to install VG for flat floor + diffuser configuration , in front leading edge of diffuser or maybe just infront where separation starts?</p>
<p><a href="https://i.stack.imgur.com/3cgTD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3cgTD.jpg" alt="enter image description here" /></a></p>
<p>Why 60° AoA, flow will separate 100% here?</p>
<p><a href="https://i.stack.imgur.com/QEiN7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QEiN7.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/NmovH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NmovH.jpg" alt="enter image description here" /></a></p>
| |aerospace-engineering|aerodynamics|car| | <p>If you mean area under the car, then there isnt much chances for the flow separation to occure. The tarmat keeps the flow connected to the car body anyway.</p>
<p>Vortex generators make sense only if the alternative is a flow separation. if flow is attached, vortex generator will increase the drag.</p>
<p>So, you put vortex generators where they are absolutely necessary, where there is enough volume for the flow to separate, and sufficiently large path has been passed by the air to accumulate turbulences that leads to flow separation.</p>
<p>Usually this means at the end of the car's bottom, as show in image, only there flow separation can occure. For top parts it could be at their middle part, as there is much more volume around to accumulate turbulences. And in case of large turns for air it might make sense to put vortex generators in the first third of the part, as flow separation becomes more likely even with little accumulated turbulences.</p>
<p>There is a way to calculate it. It usually includes distance from the start - to estimate how much turbulences each part of air has accumulated so far. And what is the hydraulic diameter - flow gets streamlined in narrow channels by itself. With significant curvature I didnt see a way to calculate it, and simulators are usually used.</p>
| 50330 | Where is best place to install vortex generators if race car diffuser stall? |
2022-03-26T14:57:17.240 | <p>I am particularly confused about finding pressure forces (specifically vertical force) on a curve surface. Although I understand the general method is to use sum of forces in the x/y direction (ie by considering the liquid weight above and so on) and find resultant force that way. I have been practising a lot of problems lately and this is one of them:</p>
<p><a href="https://i.stack.imgur.com/8VMPq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8VMPq.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/MB8ED.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MB8ED.png" alt="enter image description here" /></a></p>
<p>So here are all my attempts:</p>
<p>For the pressure diagram I did, this because pressure vary linearly.</p>
<p><a href="https://i.stack.imgur.com/WZ9pf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WZ9pf.png" alt="enter image description here" /></a></p>
<p>For the horizontal force:
<span class="math-container">$F_H=\:\frac{1}{2} \times \rho \times g \times h^2 \times b$</span>
Or in this case, <span class="math-container">$h = D = 4$</span> and <span class="math-container">$b = B = 5$</span></p>
<p><strong>Vertical force (my confusion):</strong></p>
<p>Is the sum of forces in the y-direction equals the weight of the liquid? Or does this also include the unknown <span class="math-container">$F$</span>? In other words, is <span class="math-container">$F_y = F + W$</span> or just <span class="math-container">$F_y = W$</span>?</p>
<p>Also what is the equation <span class="math-container">$x = \frac{y^2}{A}$</span> for? I am guessing you can use that to find the self weight? ie <span class="math-container">$W = mg = \rho V g = \gamma A w$</span>, in this case <span class="math-container">$A = A_2$</span></p>
<p><a href="https://i.stack.imgur.com/qRcuf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qRcuf.png" alt="enter image description here" /></a></p>
<p>for <span class="math-container">$D$</span> = 4, <span class="math-container">$x = \frac{D^2}{4}=4$</span></p>
<p><span class="math-container">$A_1 + A_2 = x \times D$</span></p>
<p><span class="math-container">$A_1$</span> can be found by integration, <span class="math-container">$A_1=\:\int _0^D\:\frac{y^2}{4}dy\:\:=\:\frac{1}{12}D^3,\:\:D\:=\:4,\:A_1\:=\:\frac{1}{12}\left(4\right)^3$</span></p>
<p>So is<span class="math-container">$F_y = W$</span> Only?</p>
<p>d) For the last part, I assume you need to make moment at the hinge? But then how can I calculate this distance from the self weight of the fluid to the hinge?</p>
| |fluid-mechanics|fluid| | <p>Hope the sketches below help the understanding of this problem. Note, the concept below holds true for any shape of the gate (straight/curved concave up/down). The difference lies in finding the volume and weight/force center of the respective shapes.</p>
<p><a href="https://i.stack.imgur.com/XffCV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XffCV.png" alt="enter image description here" /></a></p>
| 50337 | Calculating pressure forces (vertical) on a curve surface |
2022-03-27T15:49:07.990 | <p>While studying a heat exchanger, I have encountered a hydrogen rich fluid stream (enters at 90° Celsius and exits at 120° Celsius) which has composition given in the following form:</p>
<p>Air Humidity/Fluid composition: 1% (2.5 bar abs)/Mass fraction of hydrogen=0.02</p>
<p>For the heat exchanger analysis, I need to calculate several properties of this mixture like enthalpy, specific heat.</p>
<p>Property reference software like <em>Coolprop</em> and <em>Refprop</em> give data for gas mixtures, but not for air-gas mixtures.</p>
<p>How should I start to calculate the properties of this air-gas mixture?
I guess the humidity can be neglected safely because it is just 1%.</p>
| |thermodynamics|gas|air| | <p>A molar thermodynamic property <span class="math-container">$\bar{P}$</span> for a mixture containing ideal gases can generally be calculated from molar composition (mole fraction <span class="math-container">$z_j$</span>) and individual component molar property <span class="math-container">$\bar{P}_j$</span> as</p>
<p><span class="math-container">$$\bar{P} = \sum\ z_j\ \bar{P}_j$$</span></p>
<p>Composition is given. Reference values for thermodynamic molar properties for pure components abound.</p>
<p>The book The Properties of Gases and Liquids by Poling, Prausnitz, and O'Connell has information on materials properties of fluids such as thermal conductivity and viscosity.</p>
| 50348 | Properties of air and gas mixture |
2022-03-27T17:41:30.737 | <p>The mass m is suspended by a string from a disk
circular homogeneous of mass mc and radius R, as illustrated in figure 24. The disk is restricted from rotating by a spring attached to it at a distance cr of its center of rotation. If mass m undergoes a downward displacement x, determine the frequency of oscillation of the disk.</p>
<p><a href="https://i.stack.imgur.com/toAMg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/toAMg.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/QwACB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QwACB.png" alt="enter image description here" /></a></p>
<p>Can someone explain to me why m<em>x and J</em>teta are in that direction (have positive sinal) ?</p>
| |vibration| | <p>The choice of:</p>
<ul>
<li>positive axis is <strong>arbitrary</strong> (whether up/down or left/right as positive does not make a difference in the solution).</li>
<li>direction of unknown vector quantities is also <strong>arbitrary</strong>. i.e. you draw the unknown quantity as positive or negative and if the equations are properly constructed then the sign will indicate the true direction.</li>
</ul>
<hr />
<p>The most common problem in dynamics problems is what happens with derived quantities e.g. moment of an unknown force. More specifically, if it required to determine the moment of a <span class="math-container">$m\ddot{x}$</span> term (which describes an "active" force), then depending on the position of the <span class="math-container">$m\ddot{x}$</span> relative to the point that the moments are calculated the resulting moment can have a positive or negative sign.</p>
<p>An additional example --which commonly produces confusion -- is the term <span class="math-container">$J \cdot \ddot{\theta}$</span>. Although the positive direction for <span class="math-container">$\theta$</span> is defined as ClockWise(CW), the moment is positive for CounterClockWise (CCW). So in this case, I suspect what is happening is that whoever drew the diagram drew the term <span class="math-container">$J \cdot \ddot{\theta}$</span> as positive with respect to the moment positive.</p>
<hr />
<p>Apart from that I am not certain that in this problem you can ignore gravity. Which means that the force on the rope will not be equal to <span class="math-container">$m\cdot \ddot{x}$</span> (it can actually be derived by <span class="math-container">$m\cdot g -T = m\cdot \ddot{x}$</span>).</p>
<p>Also, I agree with you that <span class="math-container">$\ddot{x}$</span> should be pointing downwards, if <span class="math-container">$\theta $</span>is rotating counterclockwise (unless <span class="math-container">$\dot{\theta}$</span> is too large).</p>
<p>However, in general it will be very difficult to decypher what was the exact methodology that the person used to derive the equation (or his/her intent),without a more detailed explanation of the solution. .</p>
| 50353 | frequency of oscillation of the disk |
2022-03-27T21:06:39.267 | <p>Because I found a design for a ~80% efficient 3D printable motor, but I would like to use it as a generator from a wind turbine as well.</p>
| |electrical-engineering| | <p>Building wind turbines means that the blades need to match the generator, as power out keeps the blades from overspeeding unless you go for a offset axis machine that rotates itself out of the wind when the wind speed is too high.</p>
<p>This means you need to know the range of wind speeds that you face in your location, maximum and also the common speeds that will be the bulk of the generation.</p>
<p>One link, among many that you will find once you start to search is:</p>
<p><a href="https://www.scoraigwind.com/" rel="nofollow noreferrer">https://www.scoraigwind.com/</a></p>
<p>Met him and seen what he does....</p>
| 50358 | If I build an efficient electric motor, will I have built an efficient electric generator/alternator? |
2022-03-28T03:28:16.713 | <p>I have a marine diesel engine, and I'd like to calculate the approximate fuel consumption based on the size of the engine, at a given number of RPM - specifically for the cruising recommended cruising RPM of 2880 - 3240.</p>
<p>Guidance online varies from the super generic and unclear to PhD abstracts that are way over my head. I'm not sure exactly what details are needed but wouldn't it be the size of the cylinder (bore and stroke) x cylinder size x RPM?</p>
<p>My engine is a 3 cylinder Nanni:</p>
<ul>
<li>Max Power: 21.3 kW / 29HP</li>
<li>Compression rate 22:01</li>
<li>Bore & Stroke (mm): 78 x 78.4</li>
<li>Recommended cruising (RPM): 2880 - 3240</li>
</ul>
<p>P.S. happy to take advice using non-metric, but if so please just be precise in terms of what you're using (e.g. people talk online about "gallons" but not whether it's "US Liquid Gallon" or "Imperial").</p>
| |applied-mechanics|diesel|fuel-economy| | <p>In short, it's complicated.</p>
<p>In long:</p>
<p>What you tried is the cylinder volume times revolutions per minute. Indeed, this gives the per minute volume of air-fuel mix entering each cylinder of the engine. The point is that the air-fuel mix is not all fuel. If you can somehow get an indicative figure of the air-fuel ratio then you can calculate fuel consumption from there... but I'm not sure you can find that.</p>
<p>Engine fuel consumption is something that is measured through experiments and recorded as a function of speed (RPM) and load (torque). It is called Break Specific Fuel Consumption (BSFC) and it may be written in grams per kilowatt-hours of work done by the engine. BSFC plots are like topographic maps: small round shapes inside larger round shapes, with each shape having a constant BSFC value. And the axes of the plot are speed and load. Suppose we're looking for optimum fuel consumption. Then we need that plot, and we need to look at the smallest round shape that is inside all other, larger, round shapes. Note that it is associated with a unique speed and load combination.</p>
<p>Then, an interesting problem needs to be solved: with the said speed (RPM) and load (torque), how fast can your boat go and carrying what load!</p>
| 50363 | How to Calculate Approximate Fuel Consumption from Engine Specs |
2022-03-29T07:02:21.540 | <p>Consider that the state of stress at a point is given by an element taken at that point as shown,</p>
<p><a href="https://i.stack.imgur.com/i78Ge.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/i78Ge.png" alt="enter image description here" /></a></p>
<p>The principal stresses in the x-y plane are given as,</p>
<p><span class="math-container">$$\sigma_{1} = {\sigma_{x} + \sigma_{y} \over 2} +
\sqrt{ \left( {\sigma_{x} - \sigma_{y} \over 2} \right)^2 + \tau_{xy}^2 }$$</span></p>
<p><span class="math-container">$$\sigma_{2} = {\sigma_{x} + \sigma_{y} \over 2} -
\sqrt{ \left( {\sigma_{x} - \sigma_{y} \over 2} \right)^2 + \tau_{xy}^2 }$$</span></p>
<p>The sources that I'm referring to state that there will be another principal stress, the out of plane principal stress, which will be</p>
<p><span class="math-container">$$\sigma_3 = 0$$</span></p>
<p>The source -</p>
<p><a href="https://i.stack.imgur.com/5PWAQ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5PWAQ.jpg" alt="enter image description here" /></a></p>
<p>The book says there will be three principal planes and three principal stresses associated with those principal planes. In the second paragraph - "By making a more complete three-dimensional analysis, it can be shown that the three principal planes for a plane-stress element are the two principal planes already described plus the z face of the element", I'm not really sure about what 'complete three dimensional analysis' means, so this is how I analyzed -</p>
<p>If I take the element in the x-y, y-z and x-z plane, then for each plane (the xy, yz and xz) I would get two principal planes and two principal stresses. Which would mean there are 6 principal planes and principal stresses, in total.</p>
<p><span class="math-container">$x-y$</span> plane</p>
<ul>
<li><span class="math-container">$\sigma_{1} = {\sigma_{x} + \sigma_{y} \over 2} +
\sqrt{ \left( {\sigma_{x} - \sigma_{y} \over 2} \right)^2 + \tau_{xy}^2 }$</span></li>
<li><span class="math-container">$\sigma_{2} = {\sigma_{x} + \sigma_{y} \over 2} -
\sqrt{ \left( {\sigma_{x} - \sigma_{y} \over 2} \right)^2 + \tau_{xy}^2 }$</span></li>
</ul>
<p><a href="https://i.stack.imgur.com/8ywBT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8ywBT.jpg" alt="enter image description here" /></a></p>
<p><span class="math-container">$y-z$</span> plane</p>
<ul>
<li><span class="math-container">$\sigma_3 = \sigma_y$</span></li>
<li><span class="math-container">$\sigma_4 = 0$</span></li>
</ul>
<p><a href="https://i.stack.imgur.com/Uxnod.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Uxnod.jpg" alt="enter image description here" /></a></p>
<p><span class="math-container">$x-z$</span> plane</p>
<ul>
<li><span class="math-container">$\sigma_5 = \sigma_x$</span></li>
<li><span class="math-container">$\sigma_6 = 0$</span></li>
</ul>
<p><a href="https://i.stack.imgur.com/m1wMm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m1wMm.jpg" alt="enter image description here" /></a></p>
<p><strong>Won't there be 6 principal planes (2 in each plane xy,yz and xz) and 6 principal stresses as opposed to what the textbook says that there are three?</strong></p>
<p>Mohr's Circle-
<a href="https://i.stack.imgur.com/vjZ4R.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vjZ4R.jpg" alt="enter image description here" /></a></p>
<hr />
| |mechanical-engineering|structural-engineering|civil-engineering|structural-analysis| | <p><strong>"Won't there be 6 principal planes (2 in each plane xy,yz and xz) and 6 principal stresses...?"</strong></p>
<p>No. A plane has two faces, xy = yx, yz = zy, xz = zy, so essentially there are only three planes - one each in the direction transverse to the axes x, y, and z, as well as three principal stresses, one each projected out of the respective principal plane.</p>
<p><a href="https://i.stack.imgur.com/ImVsT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ImVsT.png" alt="enter image description here" /></a></p>
<p><strong>Mohr's circle for triaxial stresses</strong></p>
<p>Each of the circles encloses the stresses on the respective plane -
xy, yz or xz.</p>
<p><a href="https://i.stack.imgur.com/D3vgJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D3vgJ.png" alt="enter image description here" /></a></p>
| 50384 | Understanding out of plane principal stresses |
2022-03-29T17:42:19.133 | <p>using the table provided, and how it says the Velocity should be in mph, is it correct to say that the units of hp/W only taken from this table, using the formula from the table is mph also?<a href="https://i.stack.imgur.com/utnkn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/utnkn.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|aerospace-engineering| | <p>Horsepower is a measure of the amount of work done over a period of time. Specifically, <strong>one horsepower is the amount of effort required to lift 33,000 pounds one foot in one minute</strong>. Note the basic unit for velocity is <span class="math-container">$ft/s$</span> (length/time), which can be expressed as "<span class="math-container">$ft/min$</span>" and "<span class="math-container">$mph$</span>" through unit conversion.</p>
<p><span class="math-container">$\dfrac{ft}{s} = \dfrac{ft}{s} (\dfrac{60s}{1min}) = 60\dfrac{ft}{min}$</span>, and</p>
<p><span class="math-container">$\dfrac{ft}{min} = \dfrac{ft}{min} (\dfrac{1mi}{5280ft}) (\dfrac{60min}{1hr}) \approx 0.0114 mph$</span></p>
| 50399 | Are these units correct for Hp/W |
2022-03-30T10:32:31.563 | <p>Hoping to upgrade <a href="https://rads.stackoverflow.com/amzn/click/com/B07ZWQQB7D" rel="nofollow noreferrer" rel="nofollow noreferrer">this or similar telescopic door kick-down stopper</a> to release the stopper remotely. Wondering if there are standard solutions that would allow for electrical and manual loaded spring release on such telescopic kick-down door stopper using stepper motor or some other means (with kick release is preserved).</p>
<p><a href="https://i.stack.imgur.com/xE1D1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xE1D1.png" alt="enter image description here" /></a></p>
| |electrical-engineering|springs|servo|door| | <p>An electric solenoid that drops the foot when powered and releases whem unpowered seems like the best solution. This could be tied to fire systems to release doors for fire protection. Note that you run the risk here of powering the solenoid when the door is shut, and you need a solution (manual circuit break) to allow egress in a fire.</p>
<p>Electro-magnets that hold open doors to stairwells is a similar application.</p>
| 50402 | Manual/Electrically controlled loaded spring release for door kick-down stopper |
2022-03-31T11:55:59.847 | <p>Have a look at this image:</p>
<p><a href="https://i.stack.imgur.com/uUAvm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uUAvm.png" alt="enter image description here" /></a></p>
<p>this image is from: <a href="https://www.guhring.com/Tech/endmillcalc" rel="nofollow noreferrer">https://www.guhring.com/Tech/endmillcalc</a></p>
<p>I would like to know how the formula for the <strong>"Actual Chip Thickness"</strong> is done
In this case, it is 0.104</p>
<p>Using speeds and feeds for this tool:
<a href="https://www.guhring.com/ProductsServices/SizeDetails?EDP=9067370160000" rel="nofollow noreferrer">https://www.guhring.com/ProductsServices/SizeDetails?EDP=9067370160000</a>
(Using HPC-ROUGHING; Aluminium, Al-wrought alloys, Al-alloys)</p>
<p>I managed to replicate all the other formulas in my spreadsheet it's just this one I'm having an issue with. I would like to have this because it seems to affect the (actual) MRR</p>
<p>I have contacted the manufacturer and a CNC form with no luck so far
Maybe I should ask this in the Mathematics community if no one here knows?</p>
| |machining|cnc| | <p>So basically i found the answer I was looking for, I hope it helps someone else too</p>
<p><a href="https://i.stack.imgur.com/dSDMD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dSDMD.png" alt="enter image description here" /></a></p>
| 50417 | Actual Chip Thickness Calculation for CNC Machine Endmills |
2022-03-31T16:45:22.700 | <p><a href="https://i.stack.imgur.com/jJFqJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jJFqJ.jpg" alt="enter image description here" /></a></p>
<p>Say an aircraft without its engines has a mass of 118 tonnes and balances on its rear wheels only. This must mean the center of mass of the aircraft without its engines is directly above the rear wheels. If the 2 engines have a mass of 7.5 tones each and their center of gravity lies 9m ahead of the rear wheels, is it possible to find the center of mass of the whole system?</p>
<p>(If the distance between the front and rear wheels is 28.67m and the overall length of the aircraft is 66.8m)</p>
| |applied-mechanics| | <p><a href="https://i.stack.imgur.com/4BjdD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4BjdD.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$\sum V$</span> = total force = 118 + 15 = 133</p>
<p>Distance of the total force to the rear wheel (<span class="math-container">$x$</span>):</p>
<p><span class="math-container">$\sum M$</span> about the rear wheel = 0</p>
<p><span class="math-container">$x = 15*9/133 = 1.015 m$</span> - the center of the whole mass is located 1.015m to the left of the rear wheel.</p>
<p>You can verify the correctness of this solution by calculating the reactions <span class="math-container">$R_R$</span> and <span class="math-container">$R_L$</span> for both systems - 1) two loads on the beam, and 2) one (total) load on the beam, the resulting <span class="math-container">$R's$</span> should be identical.</p>
| 50418 | Center of gravity |
2022-03-31T21:34:01.110 | <p>I have a project where I need to generate a certain flow rate to levitate a ping pong ball using the Bernoulli principle. From some rough calculations, I found that it would take about 30 CFM through a 2" diameter hole to lift the ball.</p>
<p><a href="https://i.stack.imgur.com/udJEf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/udJEf.png" alt="enter image description here" /></a></p>
<p>The only fan I have is a big box fan (an axial fan) that can provide 2000 CFM. I tried taping a trash bag to my box fan and routing the flow to a small hole, but there was almost no flow through the hole. I think that I basically created an <a href="https://en.wikipedia.org/wiki/Orifice_plate" rel="nofollow noreferrer">orifice plate</a>, which requires a pressure difference to generate flow. Axial fans have notoriously low pressure ratios, so it's no surprise that this didn't work.</p>
<p>Is there any way that I could use my box fan to do what I want to do? Could I use something like a large converging nozzle on the end of my fan?</p>
<p>What if I wanted to supply multiple pipes with airflow to lift multiple ping pong balls? Could I use an axial fan? Or would something like a centrifugal fan be better?</p>
| |airflow|bernoulli|fan| | <p>Converging ducts have been constructed for years.</p>
<p>Often an exercise to work out the angle needed for lowest losses, about 6 degrees iirc.</p>
<p>Aslo an exercise to calculate the shortest joint lines especially for welders.</p>
| 50421 | Routing the flow from an axial "box" fan |
2022-04-02T23:02:56.013 | <p>It is general practice to put a bucket of water on top of a wood stove to keep air humidity under control during the burning of the stove.</p>
<p>But, such a bucket is also bound to keep some more thermal energy captured for a longer time (i.e. absorbed via the metal frame and released into the room in a more smeared-out time-interval). Does it make sense to boost such thermal "dead weight" for typical operations and given typical "family-house" stove products currently for sale on the market?</p>
<p>(Besides putting a larger water bucket on the stove, one can also add material to exposed sections of the chimney. Isn't putting heat-exchangers around chimneys common practice?)</p>
| |heating-systems|heat-exchanger|heat| | <p>Several high performance houses with lots of insulation have been designed with wood stoves that are surrounded by lots of stone to give the thermal mass. Some are still giving out heat 24 hours after the fire has gone out.</p>
<p>The name, in French speaking countries is pierre ollaire or poêles de masse.</p>
<p>This can be part of the heating design process when designing a property using passive thermal design ie using nature to its maximum.</p>
<p>We did that for a 4 bed family house that only need 2 stere or cubic metres of wood to heat annually.</p>
| 50436 | Does it make sense to increase the thermal "dead weight" of a wood stove? |
2022-04-03T19:48:41.027 | <ol>
<li><p>Can airflow through an HVAC duct be considered inviscid?</p>
</li>
<li><p>Does viscid flow ever occur for HVAC?</p>
</li>
</ol>
<p>If the answer to (1) is yes, how do we know airflow in an HVAC pipe is inviscid? In general, what's a good method for determining whether a flow is viscid or inviscid?</p>
<p>Intuitively, I associate a viscous flow with something like honey. Airflow is definitely not like honey. But I'm wondering if there's a more quantitative or theory based approach to verify that airflow would be inviscid.</p>
| |fluid-mechanics|airflow|hvac| | <p>Airflow in a duct is viscid. One of the important factors in designing ductwork for airconditioning systems is the cross-section and surface friction of the duct material.</p>
<p>Both of these two factors correlate directly with the pressure drop per lineal length of the duct and directly affect the required power of the system fan.</p>
<p>eg, cylindrical ducts are preferred over rectangular ducts because for the same cross-section area a cylinder has a smaller surface, thus less friction. Or galvanized surface has less friction with the airflow than a plastic surface.</p>
| 50447 | Can airflow through an HVAC duct be considered inviscid? Does viscid flow ever occur for HVAC? |
2022-04-04T09:58:48.993 | <p>If a glass full of water has a hole in the bottom connected to a horizontal straw with a known diameter and length. How could you apply Bernoulli to calculate the height of the glass if you could measure the flow rate at the exit and you knew the irreversible head loss of the system?</p>
| |fluid-mechanics| | <p><span class="math-container">$\dfrac{P_1}{\rho g} + \dfrac{V_1^2}{2g} + z_1$</span> = <span class="math-container">$\dfrac{P_2}{\rho g} + \dfrac{V_2^2}{2g} + z_2$</span></p>
<p>If the cup and the outlet both are open to the atmosphere, <span class="math-container">$P_1 = P_2$</span>, and with <span class="math-container">$V_1 = 0$</span>, the equation reduces to <span class="math-container">$z_1 = \dfrac{V_2^2}{2g} + z_2$</span>. Thus, the discharge velocity, at any time <span class="math-container">$t$</span>, <span class="math-container">$V_i = \sqrt{2g\Delta z}$</span>. And, the flow rate at discharge point <span class="math-container">$Q = V_ia = a\sqrt{2g\Delta z}$</span> ("<span class="math-container">$a$</span>" is the area of the outlet).</p>
<p>For the water in the cup, we can write the equation for the flow rate (a negative quantity) as,</p>
<p><span class="math-container">$Q = -A\dfrac{dz}{dt}$</span> ("<span class="math-container">$A$</span>" is the area of the cup)</p>
<p>Equate both <span class="math-container">$Q's$</span>, then integrate "z" from <span class="math-container">$z_o$</span> to <span class="math-container">$z_i$</span> and integrate t from <span class="math-container">$t_o =0$</span> to <span class="math-container">$t_i$</span>, then solve for time <span class="math-container">$t$</span>, which is the time required for the liquid to fall from <span class="math-container">$z_o$</span> to <span class="math-container">$z_i$</span>:</p>
<p><span class="math-container">$t = \dfrac{A}{a}(\sqrt{z_o} - \sqrt{z_i})\sqrt{\dfrac{2}{g}}$</span></p>
<p>Now you can find the time for draining the cup completely by setting <span class="math-container">$z_i = z_f = 0$</span>. And with <span class="math-container">$t$</span> is known, you can check the change in <span class="math-container">$z$</span>, <span class="math-container">$z_i - z_o$</span>
of any time you want quite easily.</p>
<p><a href="https://i.stack.imgur.com/9sG7o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9sG7o.png" alt="enter image description here" /></a></p>
| 50456 | Bernouli equation in a glass of water |
2022-04-04T10:17:07.007 | <p><a href="https://i.stack.imgur.com/Ua4zl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ua4zl.png" alt="enter image description here" /></a></p>
<p>Here is my attempt to the problem:
<a href="https://i.stack.imgur.com/zyuph.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zyuph.png" alt="enter image description here" /></a></p>
<p>Does this make sense? rotation clockwise 60* from horizontal to normal of the slanted plane , therefore <span class="math-container">$2 \times 60$</span> clockwise as well on Mohr's circle.</p>
<p>I am unsure why shear stress is positive downwards (as given in my textbook), also does the angle make sense? Should this angle drawn between the right triangle form or my angle on the diagram? Is the shear stress (<span class="math-container">$\tau$</span>) negative or positive? Its positive on my diagram because positive is downwards. On my teacher's solution, shear stress (<span class="math-container">$\tau$</span>) is drawn negative downwards, its shear stress is therefore negative.</p>
| |structural-engineering|structural-analysis|stresses|shear| | <p>While deriving the stress transformation equations, the angle <span class="math-container">$\theta$</span> (the angle that the outward normal of a plane makes with the positive x axis) is usually taken as positive when measured anticlockwise from the +ve x axis.</p>
<p>If we plot the shear stress as positive downwards in the Mohr's Circle, an angle <span class="math-container">$\theta$</span> in the stress element measured anticlockwise, when corresponded with an angle 2<span class="math-container">$\theta$</span> swept anticlockwise in the Mohr's Circle will give results which conform with the stress transformation equations.</p>
<p>If you consider the shear stress as negative in the downward direction, an angle <span class="math-container">$\theta$</span> measured antickw in the stress element will need to be corresponded with an angle 2<span class="math-container">$\theta$</span> swept in the Mohr's Circle in clockwise direction for the results to conform with those obtained from transformation equations.</p>
<p>So there is nothing wrong with taking the shear stress as -ve downwards, it's just that now anticlockwise angles in the stress element would correspond to clockwise angles in the Mohr's circle. In order to maintain a consistency, we take the shear stress as postive downwards, because now an antickw angle in the element
would correspond to an antickw angle in the Mohr's Circle.</p>
<p>Your teacher if draws shear stress as positive upwards will take the antickw angles on stress element as clockwise angles on Mohr's Circle and vice versa, and when he/she does that his/her results will be the same as yours - a positive shear stress on the inclined plane.</p>
<p><strong>If shear stress is taken +ve downwards:</strong></p>
<p><a href="https://i.stack.imgur.com/nHa9w.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nHa9w.jpg" alt="enter image description here" /></a></p>
<p><strong>If shear stress is taken +ve upwards:</strong></p>
<p><a href="https://i.stack.imgur.com/UiGsm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UiGsm.jpg" alt="enter image description here" /></a></p>
<p>Note: A positive shear stress in my answer means that a shear stress which tries to rotate the material in the anticlockwise direction and a negative shear stress means a shear stress which tries to rotate the material in clockwise direction. Make sure that what a +ve shear stress means to you is the same as what a +ve shear stress means to your teacher.</p>
| 50457 | How to use Mohr's circle to find stresses on a slanted plane? |
2022-04-05T04:50:36.347 | <p>I am having difficulty understanding the difference between interface capturing and interface tracking schemes. It is important to track it for solving Navier Stokes equation and other fluid problems in two-phase (multi-phase) flow. I have read a few papers (<a href="https://web.stanford.edu/%7Esjsuresh/mirjalili2017.pdf" rel="nofollow noreferrer">here</a>)</p>
| |fluid-mechanics|fluid| | <p>From my point of view, interface capturing method firstly defines a color function within on each mesh cell, that is:
<span class="math-container">$$H(x) = \begin{cases}1 &, when\; in \; fluid\;1 \newline 0 &, when\; in\;fluid\;2 \end{cases}$$</span>
and then the interface can be calculated through calculation of this color function.(For example, when VOF is applied, you need to figure out the interface zones with condition of <span class="math-container">$0 < \phi < 1 $</span>, where <span class="math-container">$\phi$</span> is the VOF function.)
While with interface tracking method, the interface is a curve(2D) or a curved surface(3D) represented by connected marker point, thus the interface is explicitly given. The image below is a sketch for interface tracking method.</p>
<p><a href="https://i.stack.imgur.com/9aBj3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9aBj3.png" alt="enter image description here" /></a></p>
<p>A useful textbook is highly recommended.</p>
<blockquote>
<p>Tryggvason, Grétar, Ruben Scardovelli, and Stéphane Zaleski. Direct numerical simulations of gas–liquid multiphase flows. Cambridge university press, 2011.</p>
</blockquote>
| 50468 | Difference between Front Tracking and Front Capturing Method |
2022-04-05T17:19:19.367 | <p>I'm looking for an elastic material (e.g., a rubber), whose tangential elastic modulus E is lower than 1 MPa (0.5 MPa would be enough). Anyone has an idea of which kind of rubber could fit this specification? I need a tube (16-25 mm diameter, and 1-4 mm wall thickness) of this material to simulate an artery (e.g., aorta) in a biomedical engineering application at room temperature. Unfortunately, arteries have a very low Young Modulus and it is very difficult to find a material with characteristics as close as possible to the ones of an artery. Thank you for any helpful information.</p>
| |elastic-modulus| | <p>Silicone gels can be this soft.</p>
<p>Your requirement is quite extreme, so you are unlikely to find a ready tube. But you can find a solution, that will harden in a required shape.</p>
<p>So, I suggest to look for liquid silicone gels, that harden when mixed with hardener, like epoxy. An example, that is probably too soft, 0.2 MPa, but gives an idea: <a href="https://www.wacker.com/h/de-se/medias/6982-EN.pdf" rel="nofollow noreferrer">https://www.wacker.com/h/de-se/medias/6982-EN.pdf</a></p>
<p>It seems tupical silicone is too tough for your use, as it is about 5 MPa, and tupical silicone gel is too soft for your use, it is about 0.01 MPa. So look for tough silicone gel. I think it is easier to make silicone gel harder, than silicone softer.</p>
| 50475 | Is there a rubber material with an elastic modulus lower than 1 MPa? |
2022-04-06T12:31:36.887 | <p>The principal stress in the textbook that I'm following is defined as</p>
<p><em>The maximum or minimum normal stress that acts on an element.</em></p>
<p>This makes sense when I take a plane state of stress and I get one maximum value and one minimum value of normal stress - <span class="math-container">$\sigma_1$</span>, <span class="math-container">$\sigma_2$</span></p>
<hr />
<p>However when we consider a 3D state of stress, there are three principal stresses - <span class="math-container">$\sigma_1$</span>, <span class="math-container">$\sigma_2$</span> and <span class="math-container">$\sigma_3$</span>.</p>
<p>Doesn't that contradict the definition of principal stress? I mean, if principal stress is the max or min normal stress then even in 3D case we should get only two values , one min and one max, how 3 values can be maximum and minimum?</p>
| |mechanical-engineering|structural-engineering|civil-engineering|structural-analysis| | <p>The definition on your textbook is almost correct. It's totally correct in <span class="math-container">$2D$</span>, but in <span class="math-container">$3D$</span> it should read something like "the principal stresses <em>include</em> the minimum and maximum normal stresses.</p>
<p>By this I mean, if you get the principal stresses in <span class="math-container">$3D$</span>, <span class="math-container">$\sigma_1,\, \sigma_2$</span> and <span class="math-container">$\sigma_3$</span> and rename them such that <span class="math-container">$\sigma_1 \geq \sigma_2 \geq \sigma_3$</span> then <span class="math-container">$\sigma_1$</span> is the maximum normal stress and <span class="math-container">$\sigma_3$</span> is the minimum normal stress, with <span class="math-container">$\sigma_2$</span> just existing there in-between.</p>
<p>Let's work out the proof of this statement. First, recall that by definition, and regardless of the coordinates used to define it, the stress tensor <span class="math-container">$\boldsymbol{\sigma}$</span> has the property of mapping a unit normal to the traction acting on the plane defined by that normal, i.e.
<span class="math-container">$$
\boldsymbol{\sigma}\mathbf{n}=\mathbf{t}
\tag{1}
$$</span>
This is illustrated in the following picture.</p>
<p><a href="https://i.stack.imgur.com/aDnmZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aDnmZ.png" alt="Traction associated with surface normal" /></a></p>
<p>The normal stress, which I'll call <span class="math-container">$\sigma_n$</span> is the projection of <span class="math-container">$\mathbf{t}$</span> on <span class="math-container">$\mathbf{n}$</span>. Let the coordinate system be defined as the principal directions, call them <span class="math-container">$\mathbf{e}_1,\,\mathbf{e}_2$</span> and <span class="math-container">$\mathbf{e}_3$</span>, under these coordinates the stress tensor is given by
<span class="math-container">$$
\tag{2}
\boldsymbol{\sigma}=\left[\begin{matrix}\sigma_1 & 0 & 0 \\0 & \sigma_2 & 0\\0 & 0 &\sigma_3 \end{matrix}\right]
$$</span>
(I'll show how to calculate the principal stresses and principal directions afterward). With this, the traction <span class="math-container">$\mathbf{t}$</span> is
<span class="math-container">$$
\mathbf{t}=\boldsymbol{\sigma}\mathbf{n}=\sigma_1n_1\mathbf{e}_1+\sigma_2n_2\mathbf{e}_2+\sigma_3n_3\mathbf{e}_3
$$</span>
and, since the normal stress is the projection (i.e. the dot product) of the traction on the normal, it's
<span class="math-container">$$
\sigma_n=\sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2.
$$</span>
Since <span class="math-container">$\mathbf{n}$</span> is a unit normal, by definition <span class="math-container">$n_1^2+n_2^2+n_3^2=1$</span>, and then
<span class="math-container">$$
\sigma_1=\sigma_1\left(n_1^2+n_2^2+n_3^2\right)\geq \sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2 = \sigma_n
$$</span>
We see that the first principal stress (which, if you recall from before, we chose to be the largest) is greater or equal to any other possible normal stress.</p>
<p>By the same argument
<span class="math-container">$$
\sigma_3=\sigma_3\left(n_1^2+n_2^2+n_3^2\right)\leq \sigma_1n_1^2+\sigma_2n_2^2+\sigma_3n_3^2 = \sigma_n
$$</span>
then the third principal stress is smaller or equal than any other possible normal stress.</p>
<p>The only step left to clarify is how to determine the principal stresses and the principal directions, so the meaning of writing the stress tensor as in <span class="math-container">$(2)$</span> is clear.</p>
<p>First, keeping in mind that the stress tensor is defined by the property written in <span class="math-container">$(1)$</span>, the principal directions are those vectors <span class="math-container">$v$</span> such that
<span class="math-container">$$
\boldsymbol{\sigma}\mathbf{v}=\lambda\mathbf{v}.
\tag{3}
$$</span>
Physically, this means that if you choose <span class="math-container">$v$</span> as the normal to the plane you are looking at, the traction on that surface is parallel to that normal. You'll immediately recognize <span class="math-container">$(3)$</span> as an eigenvalue problem. From this you can conclude that the principal directions are the eigenvectors of the stress tensor and, by extension, the principal stresses are its eigenvalues. That's all there is to it.</p>
<p>Some examples to illustrate this better. First, the most typical example from textbooks and courses, a body subject to simple shear, as shown in the figure below.</p>
<p><a href="https://i.stack.imgur.com/j4zD6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j4zD6.png" alt="Simple shear and principal stresses" /></a></p>
<p>As you'll usually see in textbooks, specially when studying Mohr's circle, a simple shear configuration, as shown, has principal stresses oriented at a 45<span class="math-container">$^\circ$</span> rotation (adapted from <a href="https://www.continuummechanics.org/principalstressesandstrains.html" rel="nofollow noreferrer">here</a>). In <span class="math-container">$2D$</span> the principal stresses would be
<span class="math-container">$$
\sigma_{max},\sigma_{min}=\frac{\sigma_{xx}+\sigma_{yy}}{2}\pm\sqrt{\frac{\sigma_{xx}-\sigma_{yy}}{2}+\sigma_{xy}^2}
$$</span>
For simple shear this reduces to <span class="math-container">$\sigma_{max}=\sigma_{xy}$</span> and <span class="math-container">$\sigma_{min}=-\sigma_{xy}$</span>. The stress tensor for simple shear is
<span class="math-container">$$
\boldsymbol{\sigma}=\left[\matrix{0 & \tau & 0\\ \tau & 0 & 0\\ 0&0&0}\right]
$$</span>
The eigenvalues of this stress tensor are <span class="math-container">$\lambda_1=\tau$</span>, <span class="math-container">$\lambda_2=0$</span> and <span class="math-container">$\lambda_3=-\tau$</span> and the eigenvectors are
<span class="math-container">$$
\mathbf{v}_1=\left[\matrix{\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\\0}\right]\qquad \mathbf{v}_2=\left[\matrix{-\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\\0}\right]\qquad \mathbf{v}_3=\left[\matrix{0\\0\\1}\right]
$$</span>
As you can see, the eigenvalues and eigenvectors correspond precisely to the results from Mohr's circle, with the unit vector in the third dimension (as all vector are eigenvectors to a <span class="math-container">$0$</span> eigenvalue.</p>
<p>Also, notice that to get this coordinate system from the canonical one you need to rotate the axes by 45<span class="math-container">$^\circ$</span> around the <span class="math-container">$z-$</span>axis, to do this you construct the rotation matrix around said axis, given by
<span class="math-container">$$
\mathbf{Q}=\left[\matrix{\text{cos}(\theta)& -\text{sin}(\theta)& 0\\\text{sin}(\theta)&\text{cos}(\theta)&0\\0&0&1} \right]
$$</span>
Rotating the stress tensor by the typical relation <span class="math-container">$\boldsymbol{\sigma}^\prime=\mathbf{Q}\boldsymbol{\sigma}\mathbf{Q}^T$</span> and you get
<span class="math-container">$$
\boldsymbol{\sigma}^{\prime}=\left[\matrix{\tau&0&0\\0&0&0\\0&0&-\tau}\right]
$$</span></p>
<p>You can apply this same procedure to other stress configuration, probably the only difference will be that <span class="math-container">$\mathbf{Q}$</span> will be a rotation in multiple axes, the rotation matrices compose as shown <a href="https://en.wikipedia.org/wiki/Rotation_matrix#General_rotations" rel="nofollow noreferrer">in the wiki</a>.</p>
| 50482 | Confusion with the definition of Principal Stress |
2022-04-06T19:50:31.627 | <p>Say we have the following indeterminate beam:</p>
<p><a href="https://i.stack.imgur.com/JgIEQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JgIEQ.png" alt="enter image description here" /></a></p>
<p>We can make it determinate by adding a pin at B, and then adding back the bending moment at B which was destroyed by the pin:</p>
<p><a href="https://i.stack.imgur.com/KVrlF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KVrlF.png" alt="enter image description here" /></a></p>
<p>My lecture handouts then go to say that we can consider the beam as two separate spans, and then use a compatibility condition to calculate <span class="math-container">$M_B$</span>:</p>
<p><a href="https://i.stack.imgur.com/SUafY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SUafY.png" alt="enter image description here" /></a></p>
<p>However, there are several things about this step that are not clear to me. How/why are we able to a split a multi-span beam into its individual spans? Is there any sort of rule that we need to follow when doing so?</p>
<p>And also, what is going on with the supports? We start with B being a pin over a roller support, and then we split into two spans, we get a roller support at B (looking at AB) and a normal pinned support at B (looking at BC).</p>
<p>I've tried to understand by drawing a free body diagram, but I don't really know how to proceed after this:
<a href="https://i.stack.imgur.com/B2K6Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B2K6Z.png" alt="enter image description here" /></a></p>
<p>Any clarity on how exactly and when we can consider a multi-span beam as separate spans would be very much appreciated.</p>
| |structural-engineering|structural-analysis|beam|structures| | <p>1 - "How/why are we able to a split a multi-span beam into its individual spans?"</p>
<p>This continuous beam constitutes a structural indeterminate system to the first degree, so there is a need to release a restrain and apply restoring force at the same location afterward to satisfy the structural compatibility (deflection or angle of rotation) conditions. In this case, the continuity of the beam at support <span class="math-container">$B$</span> was selected to be released (note that there are other ways of releasing restrains), which resulted in two independent, yet <strong>stable</strong>, beam systems.</p>
<p>2 - " And also, what is going on with the supports?...."</p>
<p>As addressed above, these two beam systems must be "stable", and the common support must be "comparable/compatible". Without in-span horizontal load, the change of roller to pin support at <span class="math-container">$B$</span> for the right side beam system <strong>has no effect on the results</strong>, only satisfying the global equilibrium requirement (a beam is unstable when supported by two rollers).</p>
<p>3 - The sketch below represents the free body diagrams after the release. It is your job to find the rotation angles at <span class="math-container">$B$</span> that satisfy the compatibility condition.</p>
<p><a href="https://i.stack.imgur.com/n7hwV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/n7hwV.png" alt="enter image description here" /></a></p>
<p><strong>Example:</strong></p>
<p>What you can say about the structural compatibility at support <span class="math-container">$B$</span>?</p>
<p><a href="https://i.stack.imgur.com/UnsPu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UnsPu.png" alt="enter image description here" /></a></p>
| 50491 | Analysing a multispan determinate beam by separating it into spans |
2022-04-07T21:36:25.347 | <p>I am having a problem determining the thermal conductivity of a layered sample. The sample is made of two materials, one whose thermal conductivity is known (λ-1) and one whose thermal conductivity is unknown (λ-2).</p>
<p>The thermal conductivity of the entire sample is known by means of testing. I am trying to calculate λ-2, and my first thought was to use the one-dimensional steady state diffusion equation without heat generation:</p>
<p><span class="math-container">$$ \frac{d}{dx}(\lambda \frac{dT}{dx}) = 0$$</span></p>
<p>Next, I will enter the known and unknown values:</p>
<p><span class="math-container">$$ \frac{d}{dx_{total}}(\lambda_{total} \frac{dT}{dx_{total}}) = \frac{d}{dx_{1}}(\lambda_{1} \frac{dT}{dx_{1}}) + \frac{d}{dx_{2}}(\lambda_{2} \frac{dT}{dx_{2}}) $$</span></p>
<p>Is this approach correct? I do not know dT (perhaps can find in machine's documentation), I'm not sure if the addition on the right side of the equation is legit; am I headed in the wrong direction?</p>
<p><a href="https://i.stack.imgur.com/EMsW0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EMsW0.png" alt="The tested layered structure" /></a></p>
<p><a href="https://i.stack.imgur.com/bOvyL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bOvyL.png" alt="A simplified model for analysis" /></a></p>
| |heat-transfer|thermal-conduction| | <p>The thermal resistance for a slab in steady-state heat transfer is <span class="math-container">$R_t = w/(kA)$</span> (K/W), where <span class="math-container">$w$</span> is thickness (m), <span class="math-container">$k$</span> thermal conductivity (W/m K), and <span class="math-container">$A$</span> cross-sectional area (m<span class="math-container">$^2$</span>). For serial thermal resistors as your case</p>
<p><span class="math-container">$$R_t = \sum R_{t,j} $$</span></p>
<p>For <span class="math-container">$N_j$</span> layers at each <span class="math-container">$k_j$</span> thermal conductivity and <span class="math-container">$w_j$</span> width, this gives</p>
<p><span class="math-container">$$R_t = \frac{N_1\ w_1 + N_2\ w_2}{k_t} = \frac{N_1\ w_1}{k_1}+ \frac{N_2\ w_2}{k_2}$$</span></p>
<p>and with <span class="math-container">$w_1 = w_2$</span> and <span class="math-container">$N_1 = N_2 = N$</span> we find</p>
<p><span class="math-container">$$R_t = \frac{2}{k_t} = \frac{1}{k_1} + \frac{1}{k_2}$$</span></p>
| 50504 | Calculate Thermal Conductivity of Layered Structure |
2022-04-08T21:37:48.833 | <p>I'm making a 1 seater vehicle. The rear axle is split into two drive axles both of which are supported on a set of a couple ball bearings mounted to the frame with housings like the following:</p>
<p><a href="https://i.stack.imgur.com/FSenf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FSenf.png" alt="enter image description here" /></a></p>
<p>I probably should have gotten roller bearings designed for axial loads but I hadn't thought of it at the time and now is a little bit too late for changes.</p>
<p>I've done a couple of test drives and things seem to be going well but it keeps bugging me - would it be possible for the bearings to pop out of the housings in the case of an extreme axial load? I can't really imagine what would cause it, but let's assume theoretically.. I'm mainly concerned because the bearings were quite easy to rotate (axially?) in their sockets prior to mounting them, I hadn't been able to take them out but having them rotate in there so easily is not very reassuring.</p>
<p>Some details that might help approximate the probability of the above occurring:</p>
<p>Vehicle maximum weight (+ driver): 200kg<br />
Drive axle length: 0.45m<br />
Drive axle diameter: 20mm<br />
Top speed: 40km/h</p>
| |automotive-engineering|bearings| | <p>Here are a few designs of ball bearings FYI. All balls are well secured and aligned securely.</p>
<p><a href="https://i.stack.imgur.com/AYaNt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AYaNt.png" alt="enter image description here" /></a><a href="https://i.stack.imgur.com/IEuDt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IEuDt.png" alt="enter image description here" /></a><a href="https://i.stack.imgur.com/cCk9P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cCk9P.png" alt="enter image description here" /></a></p>
| 50518 | Can ball bearings pop out of housings? |
2022-04-09T03:57:08.417 | <p>The following outline is inspired by something I noticed today and couldn't figure out.</p>
<p>It shows what I guess is a fairly common generic situation - an upper floor masonry partition wall supported by a beam, with further loads (ceiling/next floor joists?) supported in turn by it.</p>
<p><a href="https://i.stack.imgur.com/5bj8R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5bj8R.png" alt="enter image description here" /></a></p>
<p>I've tried to make it generic. So the supported wall <strong>(grey)</strong> is toothed into the adjacent wall <strong>(brown, left)</strong> as is common for partition walls, and the beam <strong>(red)</strong> is supported by that wall and another internal partition wall <strong>(brown, right)</strong>. Ignoring s/w and eccentricity, it's a pretty simple beam situation, where the beam supports just a wall and whatever is on the wall. I haven't given values or dimensions as I'm interested in how it's assessed, rather than some specific solution.</p>
<p>What has me puzzled is, it's not clear what governs the parts of these loads supported by the beam. There is load spreading, full-height support by the left hand wall, and the corbel angle to consider.</p>
<p>The toothed support and corbel angle suggest only a small part of the load will be supported by the beam. None of the arbitrary top loads, and maybe only a third of the wall masonry, are within the corbel angle. Can that be right? Does one reckon to include all of the topmost loads? or part of them? or none of them (and just the masonry within the corbel angle)? What is the right way to think about it?</p>
<p><strong>How can we tell which loads to include as beam loading, when loads spread throughout a wall before that wall is supported on the beam?</strong></p>
<h1>My thinking so far</h1>
<p>In theory, I've got the following points:</p>
<ol>
<li>The loads at the top of the wall will load-spread at (say) 45 degrees within the wall. They are likely to present as a UDL due to spreading, even if they began as point loads.</li>
<li>Much of the load will be taken by the left hand wall, as "our" wall is toothed into it throughout its height. (It's not like it is supported by the beam alone.) So some of the load is taken by the adjacent wall => not the beam.</li>
<li>Looking at the beam alone, the corbel angle of 60 degrees <strong>(yellow triangle)</strong> suggests the beam load includes <strong>none</strong> of the loads at the top of the wall, and only about a third of the wall's weight, because those are all ultimately outside the corbel angle and will be supported by the two masonry walls directly. Can that be right?</li>
</ol>
| |beam|bearings|masonry|load-spreading| | <p>A "corbel" is a short cantilever beam that is cast integrally with the support. The connection, through continuity, must be capable of preventing the translation and rotation displacements that cause separation of the beam and its support. In this sense, the support will take 100% of the weight of the wall and the loads on it. Lacking this continuity, the corbel will fail.</p>
<p><a href="https://i.stack.imgur.com/ws9xX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ws9xX.png" alt="enter image description here" /></a></p>
<p>On the sketch above, the red lines indicate the potential failure planes (plane of separations). Once the failure is imminent, the exact force distribution is a complicated matter; for safety reasons, it is usually assumed the support no longer effective in carrying any load, so all the loads will be transferred to the additional support system below throgh gravity.</p>
<p><a href="https://i.stack.imgur.com/BGnrI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BGnrI.png" alt="enter image description here" /></a></p>
| 50520 | Does the corbel angle govern in this scenario? If not, how is load spreading allowed for, when a load is supported by a wall on a beam |
2022-04-09T13:45:15.270 | <p>My cousin turned up a chilling report of the death of one of our relatives in the Cyfarthfa steelworks in 1905. I'm trying to form a picture of just what happened. Sorry, I only have an image of this text:
<a href="https://i.stack.imgur.com/eWhR9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eWhR9.png" alt="inquest 1" /></a>
<a href="https://i.stack.imgur.com/K1ONR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K1ONR.png" alt="inquest 2" /></a>
<a href="https://i.stack.imgur.com/0pDfS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0pDfS.png" alt="inquest 3" /></a></p>
<p>Was the "strap" a drive belt taking power from an overhead shaft? If so, how was it tangled? Can anyone imagine a picture of this? How would someone try to replace it under power, which sounds terrifying? And what would an arrangement with flange pulleys or a loose pulley look like, to make operation safer?</p>
<p>Thanks!</p>
| |safety|engineering-history| | <p>It would <em>appear</em> that in this factory, power was transmitted to individual machines by means of overhead shafting driven by a main engine where power was taken off the main shaft by means of a belt connecting a pulley on the overhead shaft with another pulley on the machine. Power was then interrupted locally by jumping the belt off the pulleys and re-engaged by guiding it back onto them, as opposed to providing some means of slacking the belt to clutch it out and tightening it to clutch it in.</p>
<p>Jumping the belt while the main shaft was turning may have been common practice at the time but is is an extraordinarily dangerous practice unless performed with consummate skill.</p>
<p>There are youtube videos of mechanics using screwdriver shanks to jump the fan drive belt onto the driven pulley of a running Volkswagen engine, a practice which should give sober minds pause.</p>
| 50524 | Historic tragedy: is this a drive belt accident? |
2022-04-10T22:48:09.773 | <p>Suppose the following beam situation within a mundane building (viewed from above). Assume vibration and torsion aren't issues. Suppose we have a choice to span a gap in 2 different ways:</p>
<p><a href="https://i.stack.imgur.com/VJTrJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VJTrJ.png" alt="enter image description here" /></a></p>
<p><em>(<strong>Grey</strong> = supporting walls, <strong>black</strong> = beam/s. Not to scale)</em></p>
<p>Clearly <strong>(B) isn't usually valid for prestressed precast lintels</strong>, unless the manufacturer advises otherwise, because of the risk of breaking over the midpoint, because the tension will be at the top, but the tensioning wires at the bottom. But if the beam is a <strong>steel profile, or perhaps timber</strong>, which handle both compression and tension fairly well, <strong>is there any common reason to prefer (A) to (B)</strong>, or any matters to take into consideration before adopting (B)? Or any reason (B) would be weaker than (A) if it used the same profile beam/s?</p>
<p><strong>Or is (B) always likely to be the obvious preference</strong> because it provides more continuity hence stiffness, has fewer connections hence freedoms/modes of failure, and is likely simpler and cheaper to fabricate?</p>
| |beam|mechanical-failure|failure-analysis| | <p>Most times, the short single span beams as shown in A are butt-jointed on the pillars to look like your diagram in B.</p>
<p>This means that the construction is easy and as a pillar is installed the next beam can be put into place to help hold it.</p>
<p>There are times a longer "beam" is used, like overhead cranes between two rails one on each side of the building.</p>
| 50537 | Are there disadvantages to using a single multispan beam, vs separate simple beams, for materials that tolerate tension as well as compression? |
2022-04-11T13:43:17.693 | <p>Say we had a material that was rigid perfectly plastic, with a stress-strain characteristic as shown:</p>
<p><a href="https://i.stack.imgur.com/tJy97.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tJy97.png" alt="enter image description here" /></a></p>
<p>Say that we are drawing this material through two dies, something like this:
<a href="https://i.stack.imgur.com/COBzI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/COBzI.png" alt="enter image description here" /></a></p>
<p><strong>What would happen if the drawing stress applied were greater than the yield stress?</strong></p>
<p>Of course this is an idealisation, and with a real material, if the stress was raised enough, the material would become plastic and then eventually fail.</p>
<p>However, what would happen with a perfectly plastic material?</p>
| |mechanical-engineering|materials|stresses|deformation|yield-point| | <p>I am trying to see what you try to say.</p>
<p>If the rate of drawing (say by creating a vacuum) is not such that it will rip the material or create bubbles or fissures it would create something like the figure below! The narrow black lines are a conceptual pattern of what will happen. My apologies for the poor sketching.</p>
<p>The necking doesn't have to be cylindrical. I tried to incorporate a bit of non-symmetry due to possible vortexes in the movement of the paste.</p>
<p>'</p>
<p><a href="https://i.stack.imgur.com/3rOp2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3rOp2.jpg" alt="plastic movement." /></a></p>
| 50545 | What would happen if we applied a stress greater than the yield stress to a rigid-perfectly plastic material? |
2022-04-12T11:23:40.747 | <p>Consider that at a particular point of the body the state of stress was that of a pure shear one, with <span class="math-container">$\tau_{xy}$</span> and <span class="math-container">$\gamma_{xy}$</span> as the shear stress and strain. I read two statements in two different books in this regard-</p>
<hr />
<ol>
<li></li>
</ol>
<p><a href="https://i.stack.imgur.com/FbvVt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FbvVt.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/uOVGh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uOVGh.jpg" alt="enter image description here" /></a></p>
<hr />
<ol start="2">
<li></li>
</ol>
<p><a href="https://i.stack.imgur.com/xpWMb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xpWMb.jpg" alt="enter image description here" /></a></p>
<hr />
<p>In 1) it is stated that there will be no change in volume of the element. In 2) it is stated that the z face will be a rhombus. I'm thinking of these statements to be corollaries of each other.</p>
<p><strong>So if I start off by saying that volume of the element doesn't change how does that makes me conclude that the x, y and z dimensions won't change as well?</strong></p>
<p>OR</p>
<p><strong>If the x, y , z dimensions do not change how can I conclude that the volume doesn't change?</strong></p>
| |mechanical-engineering|structural-engineering|civil-engineering|stresses| | <p>kamran's answer above is a very intuitive and elegant way to look at this. However, I feel it falls a little bit short when looking at the general phenomenon of shear deformation.</p>
<p>First, notice that both pictures you posted, reproduced below for clarity, as well as kamran's drawing, are bodies subjected to simple shear, meaning that there is only one of the 3 possible shear being applied.</p>
<p><a href="https://i.stack.imgur.com/Falpu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Falpu.png" alt="Simple shears" /></a></p>
<p>So it's natural to ask, ¿are all shear deformations volume-preserving? To answer this we can turn to continuum mechanics and introduce the concept of motion (in the material description) and the deformation gradient.</p>
<p>Let there be a body in a space described by the coordinates <span class="math-container">$X_1,X_2$</span> and <span class="math-container">$X_3$</span> as shown in the figure below
<a href="https://i.stack.imgur.com/R1xZj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R1xZj.png" alt="Reference configuration" /></a>.</p>
<p>The motion (or deformation) of the body is a function <span class="math-container">$\mathbf{x}(\mathbf{X},t)$</span> that describes the position of the particle originally at position <span class="math-container">$\mathbf(X)$</span> at time <span class="math-container">$t$</span>. For example, simple shear is described by
<span class="math-container">$$
x_1=X_1+kX_3 \qquad x_2=X_2 \qquad x_3=X_3,
$$</span>
which results in the following configuration</p>
<p><a href="https://i.stack.imgur.com/nZprd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nZprd.png" alt="Deformed Configuration" /></a></p>
<p>To know how much the volume changed, first we look at a line element in the reference configuration, call it <span class="math-container">$d\mathbf{X}$</span>, and we want to see how it transforms under the motion <span class="math-container">$\mathbf{x}(\mathbf{X},t)$</span>. The next figure shows the line element <span class="math-container">$d\mathbf{X}$</span> and its image on the deformed configuration, <span class="math-container">$d\mathbf{x}$</span>.</p>
<p><a href="https://i.stack.imgur.com/xYMpu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xYMpu.png" alt="Line segment transformation" /></a></p>
<p>Now we note that
<span class="math-container">$$
\frac{d\mathbf{x}}{d\mathbf{X}}=Grad(\mathbf{x}) \\
d\mathbf{x}=Grad(\mathbf{x})d\mathbf{X}
$$</span></p>
<p>But looking at the previous image, <span class="math-container">$d\mathbf{x}$</span> is equal to
<span class="math-container">$$
d\mathbf{x}=\mathbf{x}(\mathbf{X}+d\mathbf{x})-\mathbf{x}(\mathbf{X})
$$</span></p>
<p>Then, the gradient of the motion, call it <span class="math-container">$\mathbf{F}$</span>, relates how a line segment transforms under a motion. Now we can apply this to a volume element. Recall that the volume of a parallelepiped is given by the <a href="https://en.wikipedia.org/wiki/Triple_product#Scalar_triple_product" rel="nofollow noreferrer">triple scalar product</a>, then in the reference configuration the volume of an infinitesimal volume element is
<span class="math-container">$$
dV=d\mathbf{X}_1\cdot(d\mathbf{X}_2\times d\mathbf{X}_3)
$$</span><br />
and, due to the property we just established for the gradient, the volume of the volume element on the deformed configuration is
<span class="math-container">$$
\begin{aligned}
dv&=d\mathbf{x}_1\cdot(d\mathbf{x}_2\times d\mathbf{x}_3) \\
&=\mathbf{F}d\mathbf{X}_1\cdot(\mathbf{F}d\mathbf{X}_2\times \mathbf{F}d\mathbf{X}_3) \\
&=det\, \mathbf{F}\,\left(d\mathbf{X}_1\cdot(d\mathbf{X}_2\times d\mathbf{X}_3)\right)\\
&=det\,\mathbf{F}\, dV
\end{aligned}
$$</span></p>
<p>This gives us a way to evaluate the volume change for a body subjected to a given motion.</p>
<p>Going back to our simple shear motion <span class="math-container">$\mathbf{x}$</span>, it's deformation gradient is given by</p>
<p><span class="math-container">$$
\mathbf{F}=\frac{d\mathbf{x}}{d\mathbf{X}}=\left[
\begin{matrix}
1 & 0 & k\\
0 & 1 & 0\\
0 & 0 & 1
\end{matrix}
\right].
$$</span>
The determinant of <span class="math-container">$\mathbf{F}$</span> is <span class="math-container">$det\,\mathbf{F}=1$</span>, meaning a volume element will retain its volume under this transformation (because <span class="math-container">$dV=1\cdot dv$</span>).</p>
<p>If you had pure (but not simple) shear, i.e. a motion given by, for example,
<span class="math-container">$$
x_1=X_1+kX_3 \qquad x_2=X_2 \qquad x_3=X_3+kX_1,
$$</span>
then deformation gradient is
<span class="math-container">$$
\mathbf{F}=\left[
\begin{matrix}
1 & 0 & k\\
0 & 1 & 0\\
k & 0 & 1
\end{matrix}
\right].
$$</span>
Then the determinant of the deformation gradient is <span class="math-container">$det\,\mathbf{F}=1-k^2$</span>, so pure shear is not a volume-preserving transformation.</p>
| 50556 | Shear Strain and no change in volume of the element |
2022-04-13T03:24:02.223 | <p>I have a problem in finding the reaction force of the pinned-fixed structure. If I take moment at both ends, the reaction forces at A and B should be the same and the answer is wrong. How to derive the reaction forces and the bending moment?
<a href="https://i.stack.imgur.com/rBquj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rBquj.png" alt="enter image description here" /></a></p>
| |mechanical-engineering|structural-engineering|civil-engineering| | <p>A slightly more convoluted (but completely general) procedure is to work out the differential equations given by Euler-Bernoulli beam theory.</p>
<p>The first step is to replace all unknown reactions (the fixed support at A in this case) by an equivalent unknown load that will be treated sepparately. This can be done because the beam in question is considered linear elastic.</p>
<p>In this case your system becomes as follows
<a href="https://i.stack.imgur.com/4kKh7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4kKh7.png" alt="System separation" /></a></p>
<p>Now you proceed to determine the support reaction and bending moments for both these systems. Since both are simple isostatic cases I'll just state the results.</p>
<p><span class="math-container">$$
M_1(x)= \begin{cases}
\frac{P}{2}x & \text{if }x\leq\frac{L}{2}\\
\frac{P}{2}x - P(x-\frac{L}{2}) & \text{if } x> \frac{L}{2}
\end{cases}
\qquad M_2(x)=\frac{M}{L}x-M
$$</span></p>
<p>Now, for a beam with constant elastic properties <span class="math-container">$E$</span> and <span class="math-container">$I$</span> you can write the curvature as
<span class="math-container">$$
\varphi(x)=\frac{M(x)}{EI}
$$</span>
Ane Euler-Bernoulli beam theory tells us that
<span class="math-container">$$
\theta(x)=\int\varphi(x)\,dx \quad \text{and}\quad v(x)=\int\theta(x)\,dx
$$</span>
This is technically the solution to a <span class="math-container">$2^{nd}$</span> degree ODE, but it's so simple it can be solved by direct integration. Results for deflection <span class="math-container">$v(x)$</span> are stated below
<span class="math-container">$$
v_1(x)= \frac{1}{EI} \begin{cases}
\frac{Px^3}{12}+C_ax+D_a & \text{if }x\leq\frac{L}{2} \\
\frac{PLx^2}{4} - \frac{Px^3}{12}+C_bx+D_b & \text{if } x> \frac{L}{2}
\end{cases}
\qquad v_2(x)=\frac{1}{EI}\left(\frac{M}{L}\frac{x^3}{6}-M\frac{x^2}{2} +C_2x+D_2\right)
$$</span>
To find the integration constants we have the following conditions
<span class="math-container">$$
v_1(0)=0\qquad v_1(\frac{L}{2}^-)=v_1(\frac{L}{2}^+) \qquad \theta_1(\frac{L}{2}^-)=\theta_1(\frac{L}{2}^+) \qquad v_1(L)=0 \\
v_2(0)=0 \qquad v_2(L)=0
$$</span>
Solving for these constants nets you
<span class="math-container">$$
v_1(x)=\frac{1}{EI} \begin{cases}
\frac{Px^3}{12}-\frac{PL^2}{16}x & \text{if }x\leq\frac{L}{2} \\
\frac{PLx^2}{4} - \frac{Px^3}{12}-\frac{3PL^2}{16}x+\frac{PL^3}{48} & \text{if } x> \frac{L}{2}
\end{cases}
\qquad v_2(x)=\frac{1}{EI}\left(\frac{M}{L}\frac{x^3}{6}-M\frac{x^2}{2} +\frac{ML}{3}x\right)
$$</span>
With this, you need only apply the condition defining the fixed support at A. this condition is <span class="math-container">$\theta_1(0)+\theta_2(0)=0$</span>, using the value of <span class="math-container">$M$</span> to enforce this equality. Doing the algebra leads to</p>
<p><span class="math-container">$$
M=\frac{3PL}{16}
$$</span>
Exactly as the value you got on the original prompt.</p>
<p>This looks like a lot of algebra, but the main advantage is that the procedure is completely general. Following the same steps you can solve a beam having a any end support conditions, supports at the mid-span, concentrated or distributed loads, etc. Furthermore, this allows to calculate bending moment and shear at any point, which is also an important consideration. With some work you can write a program that solves this numerically, needing to work out all of this only once.</p>
<p>As a final note, if you plot the deflections <span class="math-container">$v_i(x)$</span>, positive values point downward.</p>
| 50565 | How to derive the equation for fixed-pinned beam? |
2022-04-15T11:32:29.610 | <p>The lower plastic limit theorem is stated <a href="https://en.wikipedia.org/wiki/Plastic_limit_theorems" rel="nofollow noreferrer">in Wikipedia</a> as follows:</p>
<blockquote>
<p>If an equilibrium distribution of stress can be found which balances the applied load and nowhere violates the yield criterion, the body (or bodies) will not fail, or will be just at the point of failure.</p>
</blockquote>
<p>I'm particularly interested in this theorem in the context of reinforced concrete.</p>
<p>On <a href="https://structville.com/2020/07/structural-design-of-corbels.html" rel="nofollow noreferrer">this website</a>, design of concrete corbel is described using "Strut and tie" method, which applies the lower plastic limit theorem.</p>
<p>In this method, the concrete joint is assumed to consist of struts in compression, representing parts of concrete in compression, and ties, representing reinforcing bars in tension. The struts and ties are then modeled as a truss, with members in compression and tension.</p>
<p><a href="https://i.stack.imgur.com/mfLab.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mfLab.png" alt="enter image description here" /></a></p>
<p>So, at the heart of my question is this: <strong>Do we require previous knowledge about the load paths through the structure before we can apply the lower plastic limit theorem?</strong></p>
<p>The theorem as stated on Wikipedia states that if an equilibrium distribution of stress can be found which balances the external loads which will not fail, the structure will not fail.</p>
<p>So if we first just come up with a placement of reinforcing bars, with concrete assumed to be balancing this "truss", and the rebar does not yield, nor the concrete compressive stress is not exceeded, will our structure hold (not taking into account that it can be unoptimal)?</p>
<p>Considering the picture above: we have placed a bar close to the top and assumed that the locations where we have the compression struts (concrete in compression) are locations where the concrete <em>will</em> and <em>can</em> actually be in compression. Would <em>any</em> such arrangement be correct, assuming the "truss" will not fail? Or do we need some knowledge beforehand that such arrangement is actually close to the real distribution of stresses in the structure?</p>
| |structural-engineering|civil-engineering|concrete| | <p>To answer your question: yes, absolutely any reinforcement scheme which satisfies the criterion will withstand the loading without failing.</p>
<p>Emphasis on "without failing". A poorly thought-out reinforcement layout may resist the applied load, but it will do so poorly. It won't fail according to ULS (ultimate limit state), but it won't get anywhere near satisfying the SLS (serviceability limit state).</p>
<p>So it won't collapse, but it will have cracks a mile wide.</p>
<p>Thankfully, coming up with reasonable tie-and-strut layouts is pretty straightforward, so the reinforcement you adopt will likely be somewhat near optimum.</p>
| 50588 | Question on plastic limit theorems and reinforced concrete |
2022-04-16T13:04:37.003 | <p>In an I-beam, the shear flow is usually shown drawn as on the left, not as drawn on the right. I do not understand why. I <em>do</em> understand:</p>
<ol>
<li>The total horizontal shear force must be zero, consistent with statics.This is satisfied in both images.</li>
<li>The vertical direction matches the direction of the external load.</li>
</ol>
<p>My fundamental question is, <strong>Why is the image on the left correct and the image on the right incorrect?</strong></p>
<p>It matters because when the direct shear due to a horizontal force (pushing to the right) we will subtract the shears on the upper right hand side of the left I-beam to get the net. If we were to use the right I-beam, we'd have to add the shears to get the net.</p>
<p><strong>1. How can I understand why a vertical force produces horizontal shear and
2. Why the direction is chosen as it is?</strong></p>
<p><a href="https://i.stack.imgur.com/F15i8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F15i8.png" alt="enter image description here" /></a></p>
| |beam|shear| | <p>Imagine we take an element of this beam between two cross sections of thickness dx as shown in (a).</p>
<p><a href="https://i.stack.imgur.com/liFiS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/liFiS.jpg" alt="enter image description here" /></a></p>
<p>The bending moment on one face is M and on the other is M+dM. Because the bending moments are different, at any distance y from the NA the bending stress would be different on the faces of this element.</p>
<p>Now I take a sub-element in this element (as shown in (a) by shaded part). This sub-element (see (b)) will have normal stress distributions on either side of it. However, since the stress on one of the faces is greater than the other there will be a net force on the element (see (c)). Since this sub-element should be in equilibrium there will be an internal force developed on the side face (the one which has one of the dimensions as dx). This internal force gives us a shear stress and because of the complimentary property of the shear stress we get a shear stress on the face that you are interested in, in the direction as shown in (d). You can repeat this analysis by taking an element in the lower region of the element, to get the results that you want. (Note in the lower region the stresses would be tensile).</p>
<p>As per my understanding, in the flange there will be some vertical shear stresses as well. However, in thin walled members, this vertical shear is small (in the flange) and we can assume that the vertical shear force is completely carried by the web.</p>
| 50599 | Direction of shear in beam cross section - how to understand the convention |
2022-04-17T21:27:57.303 | <p>I tried asking this on other SE forum but i think this is the correct one:</p>
<p>Basically our teacher has sent us quite a lot of exercises but I can't get past this one which says: a pulley cone consists of 3 pulleys of 150, 250 and 350 mm of diameter that link an identical but inverted cone. Determine the three possible transmission ratios. Our teacher doesn't explain very well and I don't know where to search. If someone knows something about this, anything would be helpful.</p>
<p>PD: if you think I didn't tried to solve it, in all the book there's no information about this type of mechanism and we don't see our teacher until the day we need to bring the homework. There's no more info. in the exercise sorry.</p>
<p>(ignore, i messed up the question here)Edit: by the way what I'm trying to look for here is for a formula about the transmission ratio, but I can't understand how to do the formula with this specific mechanism because I don't see how to get the speed of the pulleys, here's an example given by the book.</p>
<p><a href="https://i.stack.imgur.com/Iz4p8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Iz4p8.png" alt="example of formula about the answer" /></a></p>
<p>Edit 2: i finally came up with this answer:
This question defines a system of adjustable speed pulleys, of this type of mechanisms what we can see as a ratio is that the bigger the gear we choose the pulley will go slower, this means that if we choose the gears that are parallel to each other the speed will always be the same because we will always have gears that are proportional to each other.
350mm -> 150mm
250mm -> 250mm
150mm -> 350mm</p>
| |gears|mechanisms|homework| | <p>Your question is describing an <strong>adjustable speed pulley system</strong>. These are commonly found in bench drill presses. (See the link below.)</p>
<p><a href="https://i.stack.imgur.com/eJsx8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eJsx8.jpg" alt="enter image description here" /></a></p>
<p><em>Figure 1. A 5-speed adjustable speed pulley system. Image source: <a href="https://www.toolboxbuzz.com/power-tools/drills/skil-benchtop-drill-press-review-model-3320-01/" rel="nofollow noreferrer">Toolbox Buzz</a>.</em></p>
<p>Since the centres of the shafts are fixed and the length of the belt is fixed this arrangement requires that the path of the belt on each "level" of the system must be a constant (because the belt isn't elastic). Therefore, as you increase the pulley circumference on one shaft you must decrease it on the other to keep the path length constant.</p>
<p>Question to get you thinking:</p>
<ul>
<li>If the motor is on the left and the drill chuck on the right, will moving the belt up increase or decrease the speed of the drill?</li>
</ul>
<p>With this information you should be able to solve the problem. Update your question with the answer or method and we'll give you some feedback.</p>
| 50619 | Conical pulley system and transmission ratio (closed) |
2022-04-18T19:23:40.597 | <p>As I was installing this <a href="https://rads.stackoverflow.com/amzn/click/com/B07SHFPD8S" rel="nofollow noreferrer" rel="nofollow noreferrer">wall mount</a>, I began to wonder how exactly physics is at work with it. In other words, how does the design of the mount play a role in ensuring that the mount itself supports the television/device and distributes the forces at play? I have seen different types of wall mounts (<a href="https://rads.stackoverflow.com/amzn/click/com/B07SGD4TB4" rel="nofollow noreferrer" rel="nofollow noreferrer">like this</a>). Does the shape or structure of the mount make one much stronger (more efficient at distributing the forces) than another? Outside of allowing rotation does the swivel extension make the mount any better withstanding the forces?</p>
<p>I initially asked this question <a href="https://physics.stackexchange.com/q/704333/295975">here</a> on the Physics Stack Exchange but I think it is more suited for this platform. I came across this <a href="https://physics.stackexchange.com/a/437607/295975">answer</a> and this <a href="https://engineering.stackexchange.com/q/34878/32989">question</a> as well. Thank you.</p>
| |stresses|torque|statics|strength| | <p>Addressing the most basic but severe conditions we leave alone all the secondary connections, forces, and moments!
Let's assume:</p>
<ul>
<li>P= the weight of the TV</li>
<li>L= the length of each one of the two brackets</li>
<li>H= the height of the brackets</li>
<li>D= the distance between the two screws fastening the bracket to the wall's 2x4s or load-worthy material.</li>
</ul>
<p>The highest pullout stress on the fasteners is when the bracket is stretched at 90 degrees to the wall acting as a cantilever beam.</p>
<p><span class="math-container">$$T=C=\frac{2L*P}{D}$$</span>
Half this stress is applied at the hinge between the two sub brackets.</p>
<p>The maximum torque is when the bracket is fully extended but almost touching the wall (to pan the TV! say.). there will be a sheer force on the screws trying to rotate them off the wall.</p>
<p><span class="math-container">$$\tau= \frac{2LP}{D}$$</span></p>
<p>Intermediary torque is when the bracket is deployed in an L shape, the sub bracket coming out of the wall the second sub bracket turned 90 degrees.</p>
<p><span class="math-container">$$\tau=\frac{PL}{D}$$</span>
And the forces on the hinges are a combination of compression <span class="math-container">$ \ C=T=PL$</span> and tension and also shear due to torque <span class="math-container">$ \ \tau= \frac{PL}{H}$</span>.</p>
<p>In between angles will have the <span class="math-container">$sin(\theta)PL$</span> and <span class="math-container">$cos(\theta)PL$</span> components.</p>
<p><a href="https://i.stack.imgur.com/AWiLi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AWiLi.png" alt="tv mount" /></a></p>
| 50629 | What are the physics behind a TV wall mount? |
2022-04-19T08:19:50.607 | <p>I have been trying to wrap my head around this, but I really can't figure it out. I don't understand why the value for the specific weight of water differs when I directly convert it from <span class="math-container">$\frac{N}{m^3}$</span> to <span class="math-container">$\frac{lb_f}{ft^3}$</span>, and when I compute it using the formula for specific weight <span class="math-container">$\gamma=\rho g$</span></p>
<p>Converting it from <span class="math-container">$\frac{N}{m^3}$</span>:</p>
<p><span class="math-container">$$9810\frac{N}{m^3}\times(\frac{1m}{3.28ft})^3\times\frac{0.22lb_f}{1N}=62.4\frac{lb_f}{ft^3}$$</span></p>
<p>Using the formula <span class="math-container">$\gamma=\rho g$</span></p>
<p><span class="math-container">$$\gamma=\rho g=(62.4\frac{lb}{ft^3})(32.2\frac{ft}{s})$$</span>
<span class="math-container">$$\gamma=2009.28\frac{lb_f}{ft^3}$$</span></p>
<p>What is the cause of this? Is there something I'm missing? Something I'm misunderstanding?</p>
| |fluid-mechanics| | <p>From: <a href="https://www.engineeringtoolbox.com/water-density-specific-weight-d_595.html" rel="nofollow noreferrer">Water - Density, Specific Weight and Thermal Expansion Coefficients</a></p>
<blockquote>
<p>In the Imperial system the mass unit is the slug [sl], and is derived from the pound-force by defining it as the mass that will accelerate at <span class="math-container">$1 ft/s^2$</span> when a <span class="math-container">$1 lb_f$</span> acts upon it.</p>
</blockquote>
<blockquote>
<p>The density of water is <span class="math-container">$1.940 sl/ft^3$</span> at 39 °F (4 °C), and the specific weight in Imperial units is:</p>
</blockquote>
<p><span class="math-container">\begin{align}
γ &= \rho\ g \\
&= 1.940 sl/ft^3 \times 32.174 ft/s^2 \\
&= 1.940 lb_f/(ft/s^2\times ft^3) \times 32.174 ft/s^2 \\
&= 62.4 lb_f/ft^3\end{align}</span></p>
| 50633 | Different Values for Different Calculations of Specific Weight of Water in lbf/ft^3 |
2022-04-19T14:49:38.263 | <p>Consider a pipe in which a fluid flows such that the pipe is at a higher temperature than the fluid. At any cross section of the pipe we get a varying temperature profile. We can define a mean temperature at this cross section, which is basically the value that would give us the same energy flow rate through the cross section as the varying temperature profile would. Using this we can come up with a relation for the mean temperature <span class="math-container">$T_m$</span></p>
<p><a href="https://i.stack.imgur.com/lCGaE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lCGaE.jpg" alt="enter image description here" /></a></p>
<p>In this derivation, the enthalpy h of the fluid is substituted by <span class="math-container">$c_p T$</span>, which means this temperature <span class="math-container">$T$</span> is in Kelvins. I say this because we can write enthalpy <span class="math-container">$h= c_p T$</span> only when we have given the value of enthalpy at <span class="math-container">$T=0K $</span> as <span class="math-container">$0$</span>.</p>
<p>My question is -</p>
<p><strong>Since the temperature T is in K, does that mean I have to always remember that whenever I substitute the temperature profile (temperature variation relation) in the formula for <span class="math-container">$T_m$</span> to determine <span class="math-container">$T_m$</span> I would have to do so in kelvins?</strong></p>
<p>For instance in this practice problem they have given the temperature profile in Kelvins</p>
<p><a href="https://i.stack.imgur.com/u59mMm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/u59mMm.jpg" alt="enter image description here" /></a></p>
| |mechanical-engineering|heat-transfer|chemical-engineering|convection| | <p>Engineers work in Kelvin. Note that the question does also reference air temperature in degrees C.</p>
<p>We were often given temperatures in Kelvin, Fahrenheit, Celcius or Rankine.</p>
<p>We got in the habit of converting to K as the very first step. The most common error was using deg C and not adding 273.13...</p>
| 50640 | Mean temperature of a fluid in a pipe |
2022-04-19T18:58:25.203 | <p>I'm designing a large light art project and I've been trying to find kits that could essentially make lattice-like structures. For example how stages and cranes have lattice structures. The idea would be similar to wire frame floats for parades, but with adjustable locking joints and plastic or aluminum pipes.</p>
<p>The idea would be similar to a locking Hirth joint, but would also be universal, so I could assemble the pipes in any direction and lock them. Doesn't have to be Hirth though.
<a href="https://i.stack.imgur.com/664K7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/664K7.png" alt="" /></a></p>
<p>The final structure would be similar to this, but again fully adjustable and able to make different types of latticed structures.</p>
<p><a href="https://i.stack.imgur.com/vb1jh.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vb1jh.jpg" alt="enter image description here" /></a></p>
<p>Does anything like this exist?</p>
<p>Thanks</p>
| |joining|frame| | <p>The sketch below shows the reason why it is not possible to change the position of a locked lattice frame unless to find ways to accommodate the geometry changes at every connected joint.</p>
<p><a href="https://i.stack.imgur.com/JILx5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JILx5.png" alt="enter image description here" /></a></p>
| 50652 | Locking joints and frame construction |
2022-04-20T05:36:56.230 | <p>I have this decade old wall clock in my house and it has this little component peeping out its dial. I was curious as to what this component is and what it does in a clock.</p>
<p><a href="https://i.stack.imgur.com/6ohQ1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6ohQ1.jpg" alt="enter image description here" /></a></p>
<p>I was into Arduino for a brief period, and a photoresistor looked similar to this. Is this a photoresistor?</p>
<p>If yes, Why it's present in the clock? In 10 years I haven't seen the clock glow or light up from anywhere. What might be the reason for using the photoresistor in a wall clock like this?</p>
| |electrical-engineering|power-electronics| | <p>That is a light-dependent resistor, commonly made of cadmium sulfide. When light shines on it, its resistance goes down by a significant amount, and as such these devices found use in film cameras to control the aperture opening or shutter speed as the amount of ambient light changed- so the film would always be properly exposed.</p>
<p>They are also used widely in light fixtures, to turn them on automatically at dusk and off again at dawn. In this application they are easily interfaced to another device called a silicon-controlled rectifier which does the actual switching in response to the resistance change.</p>
<p>General Motors luxury cars from the 50's and 60's used these things in their so-called "Magic Eye" option, which would automatically kick your headlights off of high beam when the Magic Eye detected the headlights of an oncoming car.</p>
| 50660 | What is this component on the face of an analog clock? |
2022-04-20T06:46:18.710 | <p>Say I want to determine the vertical displacement of the truss at Node B</p>
<p><a href="https://i.stack.imgur.com/xHtcg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xHtcg.png" alt="enter image description here" /></a></p>
<p>How do I tell whether the vertical displacement is upwards or downwards just by looking the real forces? Where do I draw the virtual unit load? Upwards or downwards?</p>
| |structural-engineering|structural-analysis| | <p>In the image below, the left side shows the applied (horizontal) load and the reaction forces. The right side shows the virtual load in the direction of the displacement to be determined. The principal of virtual work states, that the deflection can be calculated as follows:
<span class="math-container">$$ w=\sum \frac{N_i \cdot \bar{N_i}}{EA}L $$</span>
The direction of the virtual load ist not really relevant. If the calculated deflection is positive, the deflection is in the direction of the virtual load. For a negative results it's the other way around.</p>
<p>Now, from looking at the system on the right, one can see that the only member forces are in the leftmost column (bold), i.e. those are the only two members with a non-zero virtual force. As they are in compression (virtual), while the real part (left) are in tension, you will get a negative value for <span class="math-container">$w$</span>, which means the deflection will be upwards (opposite direction of the virtual load).</p>
<p><a href="https://i.stack.imgur.com/Hu9SX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hu9SX.png" alt="Truss with reaction forces for applied load (left) and virtual load (right)." /></a></p>
| 50661 | How to know whether a vertical truss displacement is upwards or downwards (using virtual work)? |
2022-04-24T08:53:57.487 | <p>Say we have the following truss, with <span class="math-container">$1$</span> kN load pointing upwards at node 5. The goal is the calculate all the reactions and the internal forces within the members using the method of joints.</p>
<p><a href="https://i.stack.imgur.com/7zwHi.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7zwHi.png" alt="enter image description here" /></a></p>
<p>I used an online calculator to do this and got the following results:</p>
<p><a href="https://i.stack.imgur.com/kgLZo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kgLZo.png" alt="enter image description here" /></a></p>
<p>I agree with everything except the internal force in member number 4. Why is this in tension? We have a point load of <span class="math-container">$1$</span> kN in node 5, when you isolate this load and apply equilibrium equations, you have <span class="math-container">$\sum F_y = 1 + F_{4} =0$</span>, so force in member 4 is <span class="math-container">$-1$</span>. Is the software incorrect? Also And if you isolate node 4, the equilibrium equations are not satisfied?</p>
<p>Computed using <a href="https://skyciv.com/free-truss-calculator/" rel="nofollow noreferrer">https://skyciv.com/free-truss-calculator/</a></p>
| |structural-engineering|structural-analysis| | <p>In a truss, <strong>the member is in tension when the reaction is pointing away from the node</strong>. You have applied 5kN on node 5 in "+y" direction, as direct pulling, there must be an equal amount of force in member 4 but runs away from node 5 in direction of "-y".</p>
<p>Also, look at the diagram below, with the applied load and the reaction both pointing away from the respective node, will this member become longer (due to tension), or shorter (due to compression)?</p>
<p><a href="https://i.stack.imgur.com/yjyU7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yjyU7.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/DD7xF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DD7xF.png" alt="enter image description here" /></a></p>
| 50713 | Method of Joints confusion |
2022-04-24T09:19:15.303 | <p>The pictures below are screen shots of what appears to be a barn in the Ukraine. One is a top down aerial view and the other a ground up, side view.</p>
<p>What intrigues me is the purpose of the apparent "external beams" placed horizontally along the steep part of the roof. They appear to have a triangular cross section. Also, they are mounted at differing elevations. I'm guessing they have something to do with snow on the roof during winter. Does anyone know why these horizontal "external beams" are placed where they are?</p>
<p><a href="https://i.stack.imgur.com/fSXVk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fSXVk.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/FqrC3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FqrC3.png" alt="enter image description here" /></a></p>
| |building-design| | <p>In colder climates where you can expect snow and potentially freezing rain, you can get a build up of or snow or worse sheets of ice. At some point the snow or ice sheet may release it bond with the roof and begin to slide. A micro avalanche if you wish. Smooth roofs and/or those with high slopes are more prone to this effect.</p>
<p>In order to protect pedestrians in areas below the edge of the roof, it is common to put up snow guards. They appear to me to be primarily designed to break up sheets so they do not land in one solid piece on the ground but instead many smaller pieces. The staggering of the horizontal pieces could be an architectural feature or a practical one to help snow build up pass around the sides through resulting gaps.</p>
<p>Personally I would not expect the need for it on such a steep roof line for snow, but freezing rain will build up even on vertical surfaces. When the freezing rain debonds from the metal roof, these guards will help break up the sheet.</p>
| 50714 | Horizontal "external beams" on a steep section of a roof |
2022-04-25T02:03:07.037 | <p>I'm an engineering student and I've just taken my first solid mechanics course and I absolutely loved it! However, most of the problems in the textbook tend to deal with materials like steel, copper, and aluminum. I'd love to learn more about how to apply my knowledge to things which are built out of wood, where the material properties differ based on the grain's orientation.</p>
<p>I have tried looking up material properties of wood before and I've had a hard time finding straight forward answers. I was wondering if anyone would be able to point me in the right direction. Thanks!</p>
| |materials|wood| | <p>The 2018 National Design Specification (NDS) for Wood Construction is a comprehensive guide with worked-out examples.</p>
<p>This is the 2018 version. if you google NDS or AWC there are many free educational books on wood construction codes and guides available.</p>
<p><a href="https://The%202018%20National%20Design%20Specification%20(NDS)%20for%20Wood%20Construction" rel="nofollow noreferrer">NDS 2018</a></p>
| 50725 | Resources for engineering with wood |
2022-04-26T03:24:57.673 | <p>It seems like refining of oil is often done far away from where the oil was initially captured and contained from the upstream source. Why?</p>
<p>This is not an engineering question is it?</p>
| |petroleum-engineering|infrastructure| | <p>Some places, like the North Slope of Alaska, are inhospitable to the construction of refineries. It is cheaper and more convenient in this case to transport the crude to a distant refinery via a pipeline.</p>
<p>Some wells by themselves do not produce enough crude oil to justify the construction of a dedicated refinery next door. In this case, it is far cheaper to run a pipeline or railroad cars from an entire field of wells to a distant refinery, where the crude can be economically fed into the process flow along with crude from a large number of other wells.</p>
<p>Refineries require intensive infrastructure development to support their operation. So even where the weather is not a barrier and flow from the wells is large, it is cheaper to transport the crude to an existing refinery and enlarge it if necessary to handle the extra product.</p>
| 50734 | Why Not Refine Crude Oil At Source |
2022-04-26T09:49:50.060 | <p>I'm trying to get a feel for the real life behavior of hover craft air cushions and the associated power consumption. To get a feel for relevant numbers, I'd like to know</p>
<ul>
<li>actual pressure used to lift said hover craft</li>
<li>the clearance (average or range) between skirt and water/ground surface</li>
<li>the air loss through that clearance</li>
</ul>
<p>One could treat the clearance as a simple slit, however the air flow under a hover craft is more complex that that due to the round shape of the skirt. In practical operation the craft will hover around an equlibirium where the pressure is "weight of hovercraft" / "area under skirt" and the clearance is higher or lower depending on airflow (or the airflow is higher or lower depending on clearance required). An empirical formula for the relationship between pressure, clearance, skirt length and air flow would be even better, but real life numbers for one situation are ok.</p>
| |fluid-mechanics| | <p>Although I cannot assist you with explicit formulas myself, I would like to make you aware of the book <a href="https://books.google.no/books?id=aJT0gK710LwC&printsec=frontcover#v=onepage&q&f=false" rel="nofollow noreferrer"><em>Theory and Design of Air Cushion Craft</em> by Liang Yun, Alan Bliault</a>, which is available (with limitations) on Google Books. In particular, I would suggest the chapters; <em>[2] Air cushion theory</em> and <em>[12] Lift system design</em>.</p>
<p>Sure, some pages are not included in the preview, but considering the overall quality of the book, I find this a small price to pay. Having had a look at it myself, I also believe it will be of great assistance, whatever expertise you currently possess in the field of air cushion vehicles.</p>
| 50739 | What are pressures, air-flow rates and ground clearance of some real-life hovercraft? |
2022-04-26T20:34:33.867 | <p>I am looking for a solution to measure the actual total mass of carbon dioxide (CO2) emitted from a diesel engine used for power generation. Infrared CO2 sensors generally measure the gas concentration, describing the amount of gas by volume of the air, which is not useful in my application. Flow sensors are already installed in the diesel supply and return lines to determine the amount of diesel consumed, and hence the amount of CO2 created, but I need to correlate these measurements with actual measurements of CO2 mass produced at the output of the exaust system.</p>
<p>Any idea on how to approach this problem would be greatly appreciated.</p>
<p>Thanks.</p>
<p><strong>EDIT:</strong></p>
<p>The question can be reformulated as to measure the CO2 emission rate from the engine exhaust system. A typical solution in practice is to find a dedicated sensor or a monitoring system that does that.</p>
| |electrical-engineering|diesel|emissions|power-generation| | <p>Your problem isn't a trivial one; it actually falls into R&D. I disagree that the gas concentration is not a useful measurement--you could convert it into ppm/ and then into mass, even though that mass will be per volume (I'd assume <span class="math-container">$mg/m^3$</span>). It might be possible to relate that number to the mass, however.</p>
<p>In one of your comments you mention that you'd like to do it in real-time and in an industrial context--you probably will have to develop that yourself. You could start with a proof-of-concept and hook up some CO2 sensors to an Arduino or multiple Arudinos and try to get the rate of change of the concentration. If you can enclose the exhaust outlet in a closed and known volume and have the sensors be put there, that could make the development of an algorithm much easier (because you would have to develop an algorithm of how to relate whatever the sensors measure to what you want to get--and that is the mass of CO2 emitted from the diesel engine).</p>
<p>What you might want to look into is the research that has been done with various techniques to measure and <em>quantify</em> methane emissions, both in the lab and in the field. <a href="https://nap.nationalacademies.org/read/24987/chapter/5#83" rel="nofollow noreferrer">This link</a> provides some techniques which researches and people in the industry use to do so.</p>
<p>Edit: if you could do what @niel nielson has mentioned, then you'd have a baseline of what to expect. That would make the development of the above mentioned algorithm much easier, since you'd have a good estimate of which values to expect.</p>
| 50745 | Measurement of actual carbon dioxide mass emitted from a diesel engine |
2022-04-27T09:31:49.973 | <p>Let's say I have a container filled with water and its surface is in direct contact with air. How much heat/energy do I have to supply to the water such that it evaporates under its boiling point? is there even such a formula? I've looked everywhere, and the resources that I have found just calculates the amount of water (in mm/day) is evaporated, and not the amount of energy needed to evaporate the water under its boiling point.</p>
<p>Sorry if my question seems confusing.</p>
<p>Thanks.</p>
| |fluid-mechanics|thermodynamics|heat-transfer| | <p>The value you look for is the <a href="https://en.wikipedia.org/wiki/Enthalpy_of_vaporization" rel="nofollow noreferrer">Enthalpy of vaporization</a>. This is temperature dependent and can be found in a steam table (or with one of the steam table calculators)</p>
| 50750 | How do I find the amount of energy needed to evaporate water below its boiling point? |