I have already given you that data in my previous message. The uniaxial yield point is the maximum plasticity limit for the material, which is 21.1 N/mm^2.
I have an equation for polymers which describes the relationship between the shear yield point and uniaxial yield point based on the Von Mises criterion:
τy = ( ((1+μ)/sqrt(3)) /sqrt(3) ) * σy
From a diagram plotted by a stress test, I have the value for uniaxial yield point σy; the point...
Update:
The relationship between shear yield point and uniaxial yield point based on the Von Mises criterion:
τy = ((1+μ) / sqrt(3))/sqrt(3) * σy
μ is the property of the material that determines the effect of pressure on the yield point.
As a material parameter, μ is the change in yield...
But you see, you can split a force into two vectors Fx and Fy. You can do the same thing with a momentum. But my point is that you only take into consideration the force Fy which works in parallel to the cross section area. But by only looking at the given shear modulus, it is supposed to...
If a shear force is acting on a rectangular beam, how do I use the value of the shear modulus to calculate if the force applied will fracture the beam? Below I've submitted the data involved.
The shear force applied is 1000 N and the cross section area is 28 mm.
So the shear force applied is...
Material: PVC
E-modulus: 1300 N/mm^2
Poisson's ratio: 0.35
Applied shear force: 100 N
Resisting area: 28 mm^2
Shear modulus: E = 2G(1+v) --> G = E/(2(1+v)) = 1300/(2(1+0.35) = 481 N/mm^2
Does this mean that a force which can be applied to the area without fracturing is 481 N/mm^2 * 28 mm^2?
I...
lol. Are you ****ing kidding me? Torque = angular acceleration * moment of inertia.
A motor has its own lowest acceleration rating. The big gear is rigid, the small gear is not rigid.
The small gear rotates about the big gear, because the motor spins, the small gear thus also have to rotate...
NO! For every degree that it turns, the angular acceleration is changed by 0,157 rad/s^2.
So if the angular acceleration is 2 rad/s^2 at an angle of 46 degrees, then it is (2 + 0,157) at 47 degrees.
I might not have calculated ##\frac{\alpha}{\omega}##.
The rate of change of angular acceleration IS 0.15789473684210526315789473684211 rad/s^2 for every degree it turns, not radians, but degrees.
S is the distance or the arc length.
The rate of change of angular acceleration only depends on θ.
I integrated by starting with ψ instead of \alpha.
ψ = ψ
\alpha = ψt + \alpha
etc.
I am calculating the time it takes for an object to fall 90° about an axis of rotation in the x-y plane. I have managed to calculate the rate of change of angular acceleration, which will be represented by the symbol ψ.
Unfortunately it is not rad/s^3, but (rad/s^2)/θ or \alpha/θ.
The value of ψ...
This problem is making me want to tear my hair off.
I am trying to calculate the time it takes for a beam to fall to the ground.
http://myweb.lmu.edu/gvarieschi/chimney/Graph1.JPG
It would be great if I could calculate the rate of change of angular acceleration, as the acceleration would...
My solution. I think I have got it now. There seem to be ratings for nominal acceleration on stepper motors. One that I found had a nominal acceleration of 280 steps/sec, which is equal to 504 degrees / sec, which is equal to 8,79 rad / sec. Then I just need to find out whether the moment of...
I think I have come to a conclusion. In order for me to calculate whether the motor has enough holding torque rating, will be determined by its nominal speed, nominal torque etc.
So time for motor to reach nominal speed = (moment of inertia * nominal angular speed) / (nominal torque - torque...
That's a bit of a problem. Because the large gear doesn't spin. It is rigid.
Only the small gear rotate. So the small gear has to do all the work. Rotate itself, and translate itself about the axis of the large gear. As a bachelor degree student I must include as much physics as possible. But...