What would be the toque strength of a shaft? - what can be the strengths of someone
What is the assessment of the strength of the axis is the axis between the device and the load only 1 inch long and is made of stainless steel 314L is a 3 / 8 diameter rod fixed, we have the answer in PS
Saturday, February 6, 2010
What Can Be The Strengths Of Someone What Would Be The Toque Strength Of A Shaft?
1 comment:
It is a very good example of these calculations, you need http://ocw.mit.edu/NR/rdonlyres/Material ... Search the full text, 4 for example
The only difference is that we are working backwards.
First, you need the material properties. Unfortunately, the L is 314 a variety of strengths, depending on the amount of work has been hardened. I take it ignited. In this capacity, a tensile strength of 100 ksi and a yield tensile strength of 50 ksi is. (ksi 1000 psi average). But since we put the torque on the shaft without axial load, it is more convenient to use the allowable cut.
The train and shear forces are always with a ductile material by a factor of 1/sqrt (3), or 0577 assigned. Thus, the yield is in our material to 0.577 * 50 = 28.9 ksi cut.
The tensile strength for maximum power and traction have no stable relationship, but is generally advisable to incorporate a factor of 0.5 for steel. Thus, 0.5 * 100 = 50 ksi tensile strength of the court.
NUnfortunately, it is not a safety factor in your problem, choose the least among them is the performance. (You can a safety specifications in the case of the collapse at 2 a.m. to 5 p.m. to have power at the low end, for example, in this case I have the lowest of 28.9 / 2 = 14.5 ksi, and 50 / 5 = 10 ksi ).
Now we know that when the voltage reaches a part of the axis τ = 28.9 ksi to make a permanent deformation, when we are a failure. We find the pair, T on the axis that the tension generated by
τ = T * r / J
, Where R is the radius of the axis, and J is the polar moment of inertia of the cross section, equal to r ^ 4 / 2 * π. (Note that this formula gives the shear stress at any radius, the cross section. If, however, to be able to resolve the * maximum shear to the maximum value of R in the outer surface of the rod.) Thus for T.
Τ = T * J / r
T = τ * (π * r ^ 4 / 2) / r
T = τ * (π * r ^ 3) / 2
We know all about the right side, so that we can solve for T.
T = (28.9 * 1000 kg / cm ^ 2) * [π * (3 / 8 in. / 2) ^ 3] / 2
T = 300 lb-in
So take the couple on the tree can.
Now, you want an answer in the HP, but the performance is torque times the speed. So if you want an answer in the HP, you have the min rpm. To say that 500 rpm is.
Then,
T = W / W
where W is work (horses), w is the speed of rotation. So,
W = T * W
W = 300 lb-in * 500 rpm
Now it is essentially a problem of converting the unit. We know that ft-lb/min 1 hp = 33,000 by definition. So we need to change customs in feet. It is light, 12 feet = 1 must be something about these revolutions. It's a bit more complicated. 1 revolution = 2 pi radians, and a radian is a flexible dimension, so that only back. In essence, we could write 1 lap = 2 "() without unity. Now we can end the problem.
W = 300 lb-ft * (1 / 12 in.) * 500 rpm * (2 '/ 1 U)
W = 78,540 lb-ft/min * (1 ch / 33000 ft-lb/min)
W = 2.4 kW (500 rpm)
This is the answer. Of course you need to match the shaft power and speed of the engine - can also be a 1 hp motor at 1200 rpm, or a 5 min-horsepower engine to 240 rev / min or 400 hp at 3 rpm. (Good luck finding this engine, though!)
Note that the length of the shaft engage in this calculation. It is only important if you have the flexibility of the tree care - the amount of money or the amount of energy it saves.
I hope that helps!
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