How to calculate the exciting force of vibrator in kilogram
Publish: 2021-04-26 18:49:04
1. Force and weight are two different concepts and cannot be confused. In force analysis, weight (actually gravity) is often converted into force. The direction of gravity is perpendicular to the horizontal plane and points to the horizontal plane, or to the geocenter, but the direction of force is not necessarily; The point of action of gravity on other objects is the contact point of objects, but the force is the component of gravity. Only when the direction of gravity is the same as that of the component and on the same straight line, the force and gravity are equal.
2. Exciting force: FM = g / g × r ×ω 2 (G: weight of eccentric block G: acceleration of gravity R: distance between center of mass of eccentric block and rotation axis ω： Motor rotation angle (frequency)
3. The exciting force is the centrifugal force proced by the unbalanced mass of the vibration motor. The unit kn is kilo Newton. The vibration motor is also called the exciter. The parameters include model, rated exciting force, speed, rated power, rated current, installation size, installation bolt size, quality, etc. I hope I can help you..
4. Exciting force
when the centrifugal block is rotated to the highest point, the force of the plate rammer on the ground decreases<
when the centrifugal block moves to the lowest point, it provides the maximum force on the ground.
let the angular velocity of its uniform rotation be ω， If the radius is R and the mass is M1, then the centripetal force of the centrifugal block is F1, F1 = m1r ω^ 2.
the centripetal force is provided by the action of the rotating shaft on the centrifugal block F2 and the gravity m1g, F1 = f2-m1g.
therefore, the action of the centrifugal block on the plate rammer is F2 = F1 + m1g = m1r ω^ 2 + m1g.
at this time, the force acting on the ground (i.e. exciting force) is the resultant force F3 of F2 and the gravity m2 of the plate vibration rammer,
F3 = F2 + m2g = F1 + m1g + m2g = m1r ω^ 2 + m1g + m2g
it is assumed that the length of the rotating slender rod is much larger than the thickness
symbol meaning and unit
J -- moment of inertia of a rotating shaft, kg. M ^ 2
m -- mass of the revolving body, kg
I -- inertia radius, m
O -- center of gravity position< br />
when the centrifugal block is rotated to the highest point, the force of the plate rammer on the ground decreases<
when the centrifugal block moves to the lowest point, it provides the maximum force on the ground.
let the angular velocity of its uniform rotation be ω， If the radius is R and the mass is M1, then the centripetal force of the centrifugal block is F1, F1 = m1r ω^ 2.
the centripetal force is provided by the action of the rotating shaft on the centrifugal block F2 and the gravity m1g, F1 = f2-m1g.
therefore, the action of the centrifugal block on the plate rammer is F2 = F1 + m1g = m1r ω^ 2 + m1g.
at this time, the force acting on the ground (i.e. exciting force) is the resultant force F3 of F2 and the gravity m2 of the plate vibration rammer,
F3 = F2 + m2g = F1 + m1g + m2g = m1r ω^ 2 + m1g + m2g
it is assumed that the length of the rotating slender rod is much larger than the thickness
symbol meaning and unit
J -- moment of inertia of a rotating shaft, kg. M ^ 2
m -- mass of the revolving body, kg
I -- inertia radius, m
O -- center of gravity position< br />
5. F=me ω&# 178; Where m is the mass, e is the distance (length unit)
KGM / s, where M / S (i.e. mm / (s squared)) is the acceleration (because the velocity unit is m / s per second, divided by time is the acceleration); It is divided into two parts, kg is the mass, mm / S & # 178; It is acceleration, so its meaning is mass multiplied by acceleration, which is obviously force (refer to the composition of F = mg, which means that force equals mass multiplied by acceleration). In addition, considering that it is mm, we need to divide 1000 when mm is replaced by meter, and then divide 1000 when n is replaced by kn, so 1kn = 1000kg * m / S & # 178= 1000*1000kg*mm/s² That is, the sixth power of 10 kg * mm / S & # 178;
KGM / s, where M / S (i.e. mm / (s squared)) is the acceleration (because the velocity unit is m / s per second, divided by time is the acceleration); It is divided into two parts, kg is the mass, mm / S & # 178; It is acceleration, so its meaning is mass multiplied by acceleration, which is obviously force (refer to the composition of F = mg, which means that force equals mass multiplied by acceleration). In addition, considering that it is mm, we need to divide 1000 when mm is replaced by meter, and then divide 1000 when n is replaced by kn, so 1kn = 1000kg * m / S & # 178= 1000*1000kg*mm/s² That is, the sixth power of 10 kg * mm / S & # 178;
6. In the mechanical instry, the application of VSR excitation technology, we have generally named the excitation motor with the unit tonnage as the unit of measurement, only the smaller excitation motor uses the kilo Niu as the unit of measurement. And the exciter and exciting motor will not be defined by pure kn symbol, even if it is a very small unit of exciting motor, the unit is KGF, which means kilogram force, obviously, 1t = 1000kgf.
that is to say, the exciting motor with unit of 1000kg force can proce one ton exciting force at the highest level (it must be molated by variable frequency computer, and the computer is made in Qing).
that is to say, the exciting motor with unit of 1000kg force can proce one ton exciting force at the highest level (it must be molated by variable frequency computer, and the computer is made in Qing).
7. Generally divided into two categories: 1. Elastic vibrating machinery 2. Forced vibrating machinery
vibration parameters can be calculated as follows:
YM (= 0.18 / (n / 1000) 2) * (FM / ∑ g),
am = FM / ∑ g, where YM -- double amplitude, mm
n -- Vibration times, times / min
FM -- excitation force, n
∑ -- vibration weight,
am -- vibration acceleration
vibration parameters can be calculated as follows:
YM (= 0.18 / (n / 1000) 2) * (FM / ∑ g),
am = FM / ∑ g, where YM -- double amplitude, mm
n -- Vibration times, times / min
FM -- excitation force, n
∑ -- vibration weight,
am -- vibration acceleration
8. Generally, a black sign on these machines will show their vibration pressure,
their working weight is self weight plus vibration weight, and static pressure is called without vibration!
their working weight is self weight plus vibration weight, and static pressure is called without vibration!
9. Is it useful to calculate the strength of concrete after vibration
exciting force F = mw2r, where m is the weight of eccentric block and R is the eccentricity
look at these two:
http://cm.newmaker.com/art_ 10256.html
http://..com/question/29935337.html
exciting force F = mw2r, where m is the weight of eccentric block and R is the eccentricity
look at these two:
http://cm.newmaker.com/art_ 10256.html
http://..com/question/29935337.html
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