New answers tagged

6

To visualize part of Nmech's answer: in the image, the washer actually greatly increases the contact area of the bolt head. The bolt head looks pretty big: But most of that is the shaft, which obviously does not spread out load on the material. So the actual contact area looks like this: Comparatively, the bolt head on the washer looks like this: That's a ...


23

I will expand on DKNguyen answer, because to my knowledge also the two reasons are: reduce contact/bearing stresses (having a significant effect on thin finishes live galvanisation) change the joint tightening characteristics (see joint diagram). reduce contact stresses on surfaces. The basic idea is that since contact stress is defined as: $$\sigma = \...


8

Except for special applications, most washers are made of dead soft steel, which deforms under the compressive load imposed by a tightened bolt head. As the washer smooshes, it minimizes stress concentrations caused by bumps under the bolt head and surface flaws in the part the bolt is running through.


23

It is for spreading out the stress. But it is also for giving the bolt a bearing surface to turn on. The washer always goes on the side (nut or bolt) that is being turned. It prevents it from marring up the work surface and also changes the tightening characteristics. I don't know the specifics of that though but that's what I was told by a toolmaker. Always ...


1

Since this is a question regarding the "Basic Statics", let's make it simple: a) The worker will cause an axial compressive force on the ladder $F_h = Wcos\theta$. The support at the eave (ledge) should be considered as roller support that can not prevent the ladder from sliding, so the axial force must be resisted and transferred from the upper ...


0

You made two mistakes in your hand calculation: 1) Wrong equation was used to calculate the moment of inertia - you used the equation for $I_y$ instead of $I_x$, thus resulting in the hand calculation being larger than the calculation done by the FEM program ($I_y$ > $I_x$). $I_x = \frac{(9\pi^2 -64)r^4}{72\pi} \approx 0.1098r^4$ $I_y = \frac{\pi r^4}{8} ...


1

IMHO you are still confusing heat with temperature (despite the answer to this question or this ). My Q is why does increased motion or mass of molecules contribute to heat , increased temperature. I will try to answer in a way that address that fundamental difference. Consider 1 kg of water in a well insulated vessel which is travelling with 10[m/s]. ...


1

I prefer this definition for Hf: "The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states." Simply put though heat of formation is a tool we use to quickly calculate heat of ...


1

If I understood correctly from your post you started with the following cross-section (Lets say A) and you replaced it with the following (or the inverted) (let's call it B). There is one reason why you are getting larger deflections, the total second moment of area is smaller. Crosssection A The total second moment of area for Cross-section A along axis ...


0

I understood the answer to my Q. ΔH=ΔU+Δ(pV)is the general statement for Δ q and ΔH Whereas , ΔH=ΔU+pΔ(V) is a subcase when pressure is constant.


0

Von Misses Stress is a distinct state of material measure at a certain location of an element. It is not a measure of stress variations between two points. The equations below demonstrate how to calculate the Von Misses Stress for an element under various states/types of stresses. https://en.wikipedia.org/wiki/Von_Mises_yield_criterion#:~:text=Summary%20%20%...


2

IMHO (note its an opinion) in both cases the cup will fill at the same time. The reason is that the flow rate is controlled by the top vessel. Let's say it is $\dot m$. However, $$\dot m = v\cdot A$$ where: A is the cross-section v is the velocity In a unit of time, a packet of water $\Delta m$ released in time $\Delta t$ is stretched. However, the bottom ...


1

I don’t think can be achieved by gears only. The best solution depends on what your goal is, if the motor is capable of feeding al the pumps then the system will balance itself powering all the pumps with the torque they need depending on the load they are subjected to. If you need this kind of device in order to be sure one pump isn’t drawing all the power ...


3

The ball gathers velocity and when you decelerate it the inertia results in the development of forces that separate it from the cup. Basically the inertia force which is required to develop the deceleration exceeds the maximum force of the bond between the ball and the cup.


3

The weight of aluminum is very small compared to the weight of glass , presumably primarily silicate . And aluminum oxide mixes very well with silicates so there is no reason to separate them . They cannot be separated by melting . The 0.001 " thick layer of aluminum will be completely oxidized before it melts . Clearly the cost of vacuum melting is ...


1

If you only have those two nodes for the whole beam then it won't be possible, even if external loads are not applied between the nodes. The reason is the bending moments that follow (usually) a non linear relationship with x, so you won't be able to accurately interpolate between the nodes. Interpolation if more node data are available If you had more ...


1

In general, v=rω Where v is the velocity of the water, r is the radius of the turbine, and ω is the angular velocity. This is the simplest that I know of that could be quickly programmed


0

21.6% from the end, assuming the same DEFLECTION at the middle as at the end. 22.5% from the end, assuming the same STRESS at the middle as at the end.


1

UPDATE Centrifugal pumps generally obey what are known as the pump laws. These laws state that the flow rate or capacity is directly proportional to the pump speed; and the power required by the pump motor is directly proportional to the cube of the pump speed. These laws are summarized in the following equations. $$\dot V \propto n \tag{eq.1}$$ $$P \propto ...


0

Hysteresis essentially implies the loss of energy in an activity of cyclic nature.Here in case of motion of tire the portion of tread in contact with road surface is under compression due to load of the vehicle and as it moves on it gets extensional deformation. The compressed tread rubber will try to come back to its original state due to its elastic part ...


1

Basic rule of thumb would say, $$80* 3.5 =P*50 \quad P= 80*3.5/50 = 1.6*3.5= 5.6 kg$$ P is the light fixture weight but the fastners attaching the hanger to the ceiling need to be strong enough to support this load too.


0

Hookie's Law defines the linear elastic relationship of material within the elastic range in the stress-strain curve. $F = kx$ $k = EA/L$, so $F = EA(x/L) = EA\epsilon$, and $F/A = \delta$, so $\delta = E\epsilon$ Now we can generalize the "Hooke's Law" to cover the stress ($\pm\delta$), on an element of "homogeneous isotropic" material, ...


0

This problem is most easily handled using the "table" form as shown below. The result can be verified by selecting a more convenient reference line, to eliminate the potential mistake made in sign convention, as shown below. Final checks: $y_{c(A)} = y_{c(B)}$ Due to effect of substraction, the neutral axis must shifting above the centerline of ...


0

Let’s say I have a single molecule of CH4 and I want to burn this molecule. CH4 + 2 O2 -> CO2 + 2 H2O How much energy is released? Let’s say it releases 1 unit. How much of that unit will be in the form of kinetic energy? Let’s say all of it: 1 unit. How fast is that CO2 moving? How fast are those 2 H2O moving? Let’s say heat is more or less the random ...


1

Adding more wood does not change the temperature at which wood burns, but it does give you more heat. Think of it this way: a 15,000 BTU furnace may heat a small apartment, but you'll need a 100,000 BTU furnace to heat a house. They both burn natural gas at the same temperature, the latter just burns more of it at the same time. So the air coming out of ...


2

Temperature is an intensity , heat is a quantity.


1

@Ben This is to answer the comment whether there are configurations that a hydraulic cylinder acts as a spring. Please Note that with the term hydraulic cylinder, I consider any cylinder with a hydraulic fluid that can act as an actuator. Automotive shock absorber: The following is an automotive shock absorbed. Notice the use of oil and gas (usually nitrogen)...


1

Figure 1. The phase diagram for water. The pressure and temperature axes on this phase diagram of water are not drawn to constant scale in order to illustrate several important properties. Image source: Chem.LibreTexts.org. The situation you are describing is circled in Figure 1. You are travelling along the horizont line through the melting point of water ...


5

A little example: Water has a specific heat capacity (SHC) of 4.2 kJ.kg-1.K-1. I have a 2.1 kW electric kettle. $$ \Delta T = \frac {P \cdot t}{m \cdot SHC} $$ where $ \Delta T $ is the temperature rise, $P$ is power (kW), $t$ is time and $m$ is mass (kg). If I run the kettle for 60 s I will put $2.2 \times 60 = 132 \ \text{kJ} $ (energy) into the water. If ...


2

A hydraulic cylinder filled with hydraulic fluid cannot deflect unless the fluid has somewhere to go. It cannot act as a spring unless the reservoir has the capability to absorb the fluid and the ability to maintain a restorative pressure during compression. In a simple system this would be a spring and at that point you are asking, "Would a spring act ...


1

At the instantaneous moment of stopping the Container, there will be a rush of the molecules of the rear end to the front end. After some time, $\delta t$ this rush will cause extra pressure on the front of the container and less pressure on the back of the container until there is an equilibrium and the rush stops. Therefore at the time $t= \delta t$ there ...


6

In engineering terms heat (energy) and temperature are two different things. One (of many) real life examples is the kitchen stove. When you turn on the kitchen, (and leave it at a set point), the kitchen will provide thermal energy (heat) at a constant rate. However if you put a pan onto the kitchen stove you will see that its temperature rises initially ...


1

Keep adding heat to water when it is at 100 deg C causes a phase change but no more temperature change.


4

I think a review of the definition of specific heat will help: Specific heat is the amount of energy required to raise one gram of a pure substance by one degree Centigrade. $C_P = \frac {Q}{m\Delta T}$, $Q = Amount$ of $heat$ $m = Mass$ $T = Temperature$ of the $substance$ The specific heat is a constant for each specific material, and the formula can be ...


8

I agree with Chemodynamics, and I will try to add a different perspective (or maybe two). The fire process is a process where you add a fuel (i.e. energy in the system), and that fuel is gradually consumed. The rate at which it is consumed releases the chemical energy. An interesting thing here is that: the log is a three dimensional object the fire will ...


1

One basic notion behind this is that temperature of a substance is defined as the average kinetic energy of all the atoms or molecules of that substance. Now assume gas in a balloon. Also assume that the balloon is not moving, so the average velocity of the gas inside the balloon is zero. Despite the average velocity (as a vector average) being zero, the ...


19

Adding more fuel doesn't permit an arbitrarily high combustion temperature because of an unavoidable intrinsic limitation: The reaction has to heat its products. (In the case of wood, this is mostly carbon dioxide and water.) In the most efficient flame, we might mix the reactants in perfect proportion and eliminate heat losses as best we can, but we can't ...


0

Our categorization systems and atomic models are all an approximation, but they help relate the things we can't see to things we can visualize and understand. We define a chemical reaction as one where chemical "bonds" are altered. This is where electron configurations between atoms are altered. While electrons are technically indistinguishable ...


0

TL;DR: There is something wrong with the problem definition. The minimum force on the $T_{AC}$ is approximately 906 lbs at $\theta=65\deg$ (in that case $T_{AB}= 422[lbs]$) NOTE: From (Eq.3) what you really get is that: $$ \frac{T_{AC}}{ T_{AB}} =\frac{ cos (25)} {cos (θ)} $$ which actually can be interpreted, as: $T_{AC}>T_{AB}$ when $\theta > 25\...


2

Being more specific would help. Any industrial throttle or choke would have at least one valve and likely two so it can be isolated and serviced. The cooling occurs at the choke/throttle , there is no change in cross-section . In gas well heads chokes sometime have heaters to prevent freezing of hydrate ( methane + water).


1

The first thing I can see is in your $M\ddot{x}=\Sigma{F}$ equation as well as your free body diagram. You are confusing the force being exerted by the wheel with the force exerted on the robot. Free body diagrams only show the forces exerted on the body, not forces the body is exerting. The drive wheels on the robot exerts a force in the negative direction (...


3

one solution would be to spray the lubricant (provided it's fluid enough) on the gears Figure 1: source machinery lubrication By spraying you can make sure that you control the flow of the lubricant, and therefore you can control parameters like: heat abduction estimated thickness of film on the gears etc Splash lubrication Another solution is Splash ...


1

It depends on the type of leak and its geometry. If we assume the leak source is one or a few small orifices, using a more powerful pump should help raise the pressure. The orifice equation is this: $$ Q= CA\sqrt{2gh}$$ Q discharge C orifice coefficient (ranging around 80%) A area of the orifice h is the head Therefore, since the extra leak is ...


0

Because 3d printing a hard case solution, will create a hard barrier which will affect the weight measurement, you'll probably need to consider some sort of flexible isolation method. A quick (and not so eye pleasing solution) if a) the project does not need to be submerged in water b) if you are not so much bothered about appearances is to consider ...


7

In the ones pictured, there are no trailing axles - they are a 2-6-0 and a "tenwheeler" 4-6-0 respectively. The ones shown have the firebox between the two rear drive axles. .That was an evolution of the earlier 4-4-0 arrangement. In later 4-6-0, "Prairie" 2-6-2 and "pacific" 4-6-2s and tank engines with those configurations, ...


0

The procedure below demonstrates a structural engineering approach: Determine the geometries of the rod and the clamp, and find the gap. Through geometric compatibility, determine the force ($F_B) required to close the gap using equations for the curved beam. By definition, $T = F*a/2R$, which is to be resisted by the friction force ($F_f$). The friction ...


0

The simplest approach, I think, will be using two stacked gears on a single shaft - one with 2/3 of its teeth missing, one with 1/3 of them missing. The gear with 1/3 gears missing engages to output gear (2/3 of the time) through two gears, the one with 1/3 missing - through one gear, spinning freely while the other one is engaged and vice versa. This way ...


1

The only problem I see with your calculation (some might argue against it) is that you are calculating the screws in shear. IMHO, the best practice is design the bolt, so that the axial force of the force is such so that the friction force between the two plates will transfer the torque between the machine elements. So instead of $F_{tau}$, I would calculate ...


0

Typically, a pass-thru for a fluid conduit going through a hole in a wall is called a bulkhead fitting. It's basically a tube with fitments on both ends to accept tubing connectors, and large nuts that screw onto the ends of the tube which, when tightened, "capture" the wall on both sides and hold the bulkhead fitting firmly in position.


1

Very short answer: The tension in the cable must be greater than the weight since it's supporting both the weight and the centripetal force. Cosines are always less than 1.


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