New answers tagged mechanical-engineering
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If you mean a cylinder with the diameter $Dcm$ and a square with a side $A=Dcm,$ The square is stronger, solid, or hollow.
If they have the same moment of inertia, I, then they are equally strong only for bending moment, but then the square post has a bit less area and a bit weaker for the axial load. let's see what size square has equal I as a cylinder with ...
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If by "steady state" you mean the output shaft isn't rotating, you are correct if the applied torques cancel out.
For example, if the sun shaft is your desired output with the ring gear and carrier as the inputs, rotating in the same direction, the net torque will be (Tr x Ns/Nr)-(Tc x Ns/(Nr+Ns)). The output torque is the sum of the input torques ...
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Use this fork:
https://github.com/AeroDME/NASTRAN-95
Use gFortan and cmake to build it on windows 10.
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I am still not certain 100% what you are trying to do, but from what I understand the wo solution you might want to consider are
chains/pulleys : if the position of the shafts is fixed
Schmidt linkage ( Video for operation ): if you want to move parallel one axis of rotation with the others.
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Put teeth on the circumference of the driver and then a gear on the shaft of the driven. Either they can mesh or use a chain or belt drive - depends on the torque/speed/power...
Of course if space is an issue consider a hydraulic link but is the speed ratio to be held constant or is there an acceptable tolerance.
Another option is to put gears on both shafts ...
1
I found this definition quite interesting, you can comment on it.
"Centripetal acceleration is defined as the property of the motion of an object, traversing a circular path. Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration."
Better picture.
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Your nomenclature and variable name choices are a horror. I'll try and redo your stuff in the canonical way
Rotational velocity $\displaystyle = \omega = \frac{s}{r\cdot t} = \frac{\theta}{t} \
$seconds$^{-1} $
Acceleration has the units of meters per second$\,^2$, the formula for centripetal acceleration is $\omega^2\cdot r$
In steady rotation this is ...
1
Since we are working on force vectors. You can should always verify your answers graphically with direction in mind. The diagram below is an example utilizing the original sketch and using parallelogram method. Note that in which the vectors (500N & 800N) shall be drawn to correct length and direction, thus the resultant R can be directly measured off
...
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50 and 22 refer to diameters of the conical bit. Usually in axisymmetric objects you denote diameters. It is the same as the 36 and 45 diameters (which are more easily understood).
So it is:
$$\sqrt{\left(\frac{50-22}{2}\right)^2 +14^2}$$
$$\sqrt{\left(14\right)^2 +14^2}$$
So there is no contradiction.
Regarding the 15, it actually refers to the distance ...
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Here's my solution in GNU Octave. Everything is in inches. The formula you used (fix-fix support) gives
0.412" deflection for S-40 and 0.437" deflection for S80.
clc, clear
format long
E = 400000 ;
D_1 = 1.315;
d_1 = 1.049;
d_2 = 0.957;
w_1 = 0.32/12;
w_2 = 0.41/12;
L = 10*12;
I_1 = pi/64 * (D_1^4 - d_1^4);
I_2 = pi/64 * (D_1^4 - d_2^4);
fprintf(...
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The problem in your solution is a confusion about angle $\theta$ from the cosine rule you are using. The triangle of forces will look like the following image:
where r is the angle in the cosine rule:
$$R^2 = F_1^2 +F_2^2 - 2F_1 F_2 \cos(r)$$
$$R^2 = F_1^2 +F_2^2 - 2F_1 F_2 \cos(180-\theta)$$
If this is superimposed to your initial image it will look like ...
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In order to have a successful design, the beam needs to satisfy the following conditions:
fv <= fv' (shear), fb <= fb' (bending) , and y_max <= y_permissible (deflection)
fv' and fb' are allowable stress of shear and bending, respectively. For ductile metal, fv' is usually limited to 0.4Fy, and fb' is in the range of 0.6Fy - 0.66Fy. The permissible ...
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For schedule 40 pipe:
I = \Pi x (1.315^4 - 1.049^4)/64 = 0.0873 in^4
\delta = [0.32(lb/ft) x (10 ft)^4 x (12 in/ft)^3]/[384 x (400000 lb/in^2) x (0.0873 in^4)] = 0.4124"
Please check my math.
1
The formula is correct, the answer is likely wrong.
Let's look at the two variables that can affect the outcome.
The weight and the I.
The schedule 80 I is.
$$I = π (r_o^4 − r_i^4) / 4 = \pi ( 1.315^4-0.975^4)/4= 1.898"^4$$
$$\delta= \omega*120"^4/384EI $$
The schedule 40 I is.
$$I = π (r_o^4 − r_i^4) / 4 = \pi ( 1.315^4-1.049^4)/4=1.397"^4$$
this gives a 1....
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You probably have mixed up the units (i.e. you have 10 ft pipe and inches for the Moment of area)
The formulas I've used:
$$ d_{max}= \frac{5wL^4}{384EI} $$
$$I = \frac{\pi (d_o^4 − d_i^4)}{ 64}$$
the values I've used are the following:
Type
Outer Diameter
Inner Diameter
Weight
Schedule 40
1.315” (0.0334m)
1.049” (=0.0266m)
.32 lbs/ft (=4.67 N/m)
...
0
If you shut off the ignition for the cylinder you wish to disable, the action of the piston rising and falling in the cylinder (with no combustion) will tend to load the combustion chamber with motor oil, which then fouls the spark plug- so that when you want to turn that cylinder back on, it refuses to fire.
That puts the driver in a perilous situation, if (...
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The general equation of an object moving uphill is FT >= FR, in which FT is the thrust produced by the object, and FR is the resistance force that impede the movement of the object. FR includes,
Downward pull of gravity, F1 = mgsin(\theta), the angle \theta = slope (10 degrees)
Friction of tires on pavement, F2 = mgcos(\theta)(\mu), \mu is the friction ...
2
The simplest idea for me would be to add on the bottom (and the top cover) small inserts of the following shape (dimensions will vary depending on the size of the ball, etc)
They would guide the ball towards the opening with the right velocity.
You only need to provide adequate clearance for the rotating part, which from your side view should not be a ...
3
Figure 1. The cylinder is 24 mm diameter with an outer thread of M24 x 2 (pitch) extending to 16 mm from the end.
And what is the diameter of the rotor? Is it 40? if yes then why they didn't put any diamater mark next to the number 40?
Yes, 40 mm. It should be marked with a Ø symbol or "DIA".
And I cannot find out the angle of the central key ...
1
There are a few things that you need to consider:
10% road inclination
the force from 10% road inclination would be:
$$F_{incl} = m\cdot g \sin(\phi)\approx 5313[N]$$
where:
m= 12000[lb]
$\phi=tan(10\%)= 5.7[deg]$
Air Drag
The air drag is given by:
$$F_D= \frac{1}{2} C_D \rho_{air} A v^2\approx 3200$$
where:
$C_D =0.608 $ coefficient of Drag (value is ...
1
Photons either pass through a given material or they don't.
If they pass through, it's as if the material isn't there as far as they're concerned (ignoring refraction).
If they get absorbed, then they dump all their energy into the material. This might excite the incident atom such that it then re-emits the photon with the same or different wavelength, such ...
1
I've found this method on YouTube (Channel: thang010146), but as NMech shows there are a few other methods like the lead screw and the worm drive screw jack.
But all look like they work with the same principle of using the friction of the thread to stop sliding. So we just have to match the friction to the max load capacity we want.
3
There are many ways this can be achieved, and it will be depended on the actual implementation.
One very prominent example is the lead (or helix) angle in lead/power/acme screws.
If you do the analysis you will end up with the following
[
where:
T = torque
F = load on the screw
dm = mean diameter
$\mu$ = coefficient of friction
l = lead
$\phi$ angle of ...
0
Unfortunately the answer is not a driven cable drum with dynamic breaking. I suggest you think on a more mundane braking system. Possibly harnessing the ropes own weight to create the tension would be more feasible with the length and size of rope you are using.
To explain why I believe the use of dynamic breaking is not viable I have outlined the problems ...
0
In compression, except for very short members the member fails by buckling under Euler's formula:
$$P_c= n\frac{ π^2 E I}{ L^2} $$
Where n is a factor based on the column's end connections.
Therefore we see that with the same area of a section if we can increase its I, second area moment, we increase the compression force the member can safely take.
The I ...
0
You are correct on both counts.
For members under tension, both cross-sections should behave equally, since the limit to the allowable tension force is simply:
$$P = f_y A$$
Where $f_y$ is the material's yield strength and $A$ is its area. So if both cross-sections have the same material and area, they'll resist the same load.
However, compression is ...
0
There is a fundamental difference between tension and compression. Bars under compression are prone to buckling.
Compared to tensile strength which depends on the cross-sectional area A (Compressive strength also depends on A), buckling of compressed members depends on
the length L
the modulus of elasticity E
the second moment of area I
The second moment ...
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The young's modulus (E) is the best indicator of material strength. In terms of rigidness and flexibleness of a structure of certain material, you need to relate E to the deformation behavior and geometric properties. In general, for members subject to concentric axial force (tension/compression), the rigidity can be expressed as L/EA, and the flexibility is ...
2
Another way to approach is to determine the properties that influence the specific problem at hand. I'll give an example and you can see if it works for you.
Define the problem.
Lets assume that you want to determine the most appropriate material for the tensile testing of a bar of length L (with a uniform cross-section), and safety factor N, so that the ...
0
Regeneration is possible in DC Series Motor since the field current cannot be made greater than the armature current. Since the resistance of the field winding is low, a series resistance is connected in the field circuit to limit the current within the safe value.
1
In flow problems, it's generally a good idea to work backwards in realtion to the flow: Find where each branch is ending and the pressure at this point (height, atmospheric pressure, $p_{out 1}, p_{out 2}$ ). Then, find the relation between pressure loss and flowrate for each branch ($\Delta p_1(Q_1), \Delta p_2(Q_2)$).
You know the the pressure at the ...
2
There a few steps in this problem.
The first one is to figure out what is the torque that each shaft is subjected.
The second one is to determine the twisting angle.
In this you are making the assumption that the disks do not deform.
So the regarding the torque (if you know the basics about gear assemblies this is obvious):
the short rod (L/2) is ...
1
A problem with meaningful solution must have a key with enough conditions/constraints, that make it soluble. As a problem solver, we need to find/identify the key and the associated conditions/constraints. Sometimes work a problem in the backward manner can be helpful. The steps are as follows:
Expand the unknown (R = ?) to a more detailed expression. That ...
2
I don't have time to type this up nicely - thought it would be quicker to write out by hand but in hindsight my handwriting is terrible. Let me know if you can't read anything!
3
In order for the Resultant force to be vertical you only need the sum of the horizontal components of $F_1$ and $F_2$ to be zero
$$F_{1x} - F_{2x} =0 $$
$$F_{1}\cos(70) - F_{2}\cos\theta =0 $$
$$\cos\theta=\frac{F_{1}}{F_{2}}\cos(70) - F_{2} $$
$$\theta=\arccos\left(\frac{F_{1}}{F_{2}}\cos(70) - F_{2}\right) $$
if you substitute you should get
$$\theta=50[...
0
Go outside start your car and rev it up you will feel it pull to the right and it's all about the load. Look around you'll see the same trucks racing without trailers and they spend the tires. Once you add that much weight all that spinning has to go somewhere I don't care how strong your frame is it's going twist watch any kind of high horsepower drag ...
0
"Transverse" is quite clear, it is an axis that makes 90 degrees to the reference axis. So, if the reference axis is x-axis, both the y and z axes are "transverse" to the x-axis.
"Lateral" has a few different meanings and usages:
a) "The lateral distance...", means the horizontal distance between the referenced axis ...
0
High speed steel cutters dull easier because they are less resistant to heat than a carbide tool. Low feed rates tend to increase heat build up in the tool for the same amount of cutting.
Your suggestion to reduce transverse loads by slowing the feed rate will help with accuracy, but continues to run the risk of excessively heating the tool. All other things ...
0
Rotational speed is a by-product of the desired result and is generally a compromise to optimise other factors.
To give a simple example of why this is, consider that a twist drill cuts across its entire frontal surface. While the rotational speed of the bit is constant across the face, the linear speed varies from 0 (at the centre) through angular rotation ...
0
If describing a material property such a steel ; lateral would never ,correctly, be substituted for transverse.
3
Mill hardening is a catch-all term for any set of processes that produce a standard alloy with defined properties from the metal mill. It is used when talking about treatments by a material vendor that a workpiece will have been received with. In the shop you may apply further treatments to different sections, but the "mill hardening" would refer ...
5
Mill hardening would refer to work hardening, where cold-rolling the material distorts the grain structure and causes dislocation pile-up and entanglements, which reduce the material's ductility and thereby increase its yield strength.
0
You have correctly recognized it is the buckling phenomenon that causing the member to fail before reaches its tensile capacity, which means, due to slenderness effect, the condition fcr < fc < fy has occurred. In which fc is the compressive stress of the member, fcr is the limiting Euler Buckling Stress, fcr = \PiEI/(KL)^2, KL is the effective (...
5
Normally fans are not custom built by a company that does not build fans. The concept of a centrifugal fan is not particularly complicated, but it is difficult to achieve target flow rate, pressure, and efficiency if your company does not do it on a daily basis. If you are not able to purchase a fan, another option is to copy someone else's simple straight ...
1
On the motor spec sheet you are correct to used the "Gear ratio 1/300" graph because that is the motor+gearbox that you intend to purchase. Those other graphs are for different models that the company offers for sale.
I did not have time to dive into all your previous questions. So I will just explain what this motor can do and you can decide if it ...
1
The AC power plugs vary a lot country to country. You will need to know how you plan to attach the pins to your circuit; ex wire solder cup, wire crimp, pcb surface mount solder, or pcb through hole solder. The exact manufacturing methods vary greatly based on those choices.
Unless you have unique requirements, I would definitely recommend sourcing pins ...
0
A transmission system influences the torque of an electric motor and is required to improve or to advance during certain tasks performed while the motor is used in carrying load.
Nowadays, we have electric vehicles and they are used to carry the load of the whole car from one place to another.
According to Ohm's law, V=IR relation, when the load is applied, ...
0
Assuming a frictionless roll of the 60 ton the horizontal force pushing the mass up the ramp is. Or if there is friction we need the data on that.
$\theta= slope\ angle$
$\alpha = acceleration$
We care about acceleration not the speed of the object because a constant acceleration will push the object to any speed in a sufficient time.
$$F_h= 600000kg*9.8* ...
0
You are trying to transform a 3 loads system (P system) to an equivalent 4 loads system (F system) that has identical pressure diagram below the disc. Assume the disc is a rigid body, it can be easily done by setting up the coordinate system, and solving by satisfying the equilibrium conditions [sum F_z, sum M_x, sum M_y]_P = [sum F_z, sum M_x, sum M_y]_F.
...
1
The short answer is - always verify power AND torque for any system. Then you never have any surprises.
Pete's guess on speed (1000 rpm) is in the ballpark. Most hydraulic gear pumps/motors have a minimum operating speed around 300-500 rpm. I'll use 800 rpm as a conservative example. At that speed your flow rate would be
Q = (0.25 in^3/rev) * (800 rev/min) / ...
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