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Analysis of chain drive operation

analysis of chain drive movement

1. Uneven movement of chain drive

a broken line is formed after the chain enters the sprocket, so the movement of chain drive is very similar to that of belt drive around regular polygon wheels, as shown in Figure 1. The side length is equivalent to the chain pitch P, and the number of sides is equivalent to the number of sprocket teeth Z. Each revolution of the sprocket, the distance the chain moves is ZP. Set Z1 and Z2 as the number of teeth of the two sprockets, P as the pitch (mm), and N1 and N2 as the speed of the two sprockets (r/min), Then the average speed of the chain V (m/s) is v=z1pn1/60*1000=z2pn2/60*100 "with the weight of most new aircraft more than half from composite materials 0

the average transmission ratio of the chain transmission can be obtained from the above formula i=n1/n2=z2/z1

in fact, the instantaneous chain speed and instantaneous transmission ratio of the chain transmission are changing. The analysis is as follows: set the tight edge of the chain to be in a horizontal position during transmission, see Figure 1. Set the driving wheel to be at an equal angular speed ω 1. Rotation, high test repeatability; If the latter is cheap, its indexing circumferential speed is R1 ω 1。 When the chain link enters the driving wheel, its pin shaft always changes its position with the rotation of the sprocket. When in β The instantaneous speed of the horizontal movement of the chain is equal to the horizontal component of the circumferential speed of the pin shaft. That is, chain speed vv=cos β R1 ω 1

the variation range of angle is ± φ Between 1/2, φ 1=360。/z1。 When β= At 0, the chain speed is the maximum, vmax=r1 ω 1； When β= ± φ At 1/2, the chain speed is the minimum, vmin=r1 ω 1cos( φ 1/2) 。 Therefore, even when the driving sprocket rotates at a uniform speed, the chain speed V changes. The chain pitch changes periodically every time it turns, as shown in Figure 2. Similarly, the instantaneous speed of the vertical movement of the chain v`=r1 ω 1sin β It also makes periodic changes to make the chain shake up and down

Figure 2 chain speed change law

driven sprocket due to chain speed v ≠ constant and γ The constant change of angle (Figure 1), so its angular velocity ω 2=v/R2cos γ It is also changing

The instantaneous transmission ratio I of the chain transmission ratio is I= ω 1/ω 2=R2cos γ/R1cos βobviously, the instantaneous transmission ratio cannot obtain a constant value. Therefore, the chain drive is unstable

2. Dynamic load of chain drive

the main reasons for the dynamic load of chain drive during operation are:

(1) the chain speed and the angular speed of driven sprocket change periodically, resulting in additional dynamic load. The greater the acceleration of the chain, the greater the dynamic load. The acceleration of the chain is

. It can be seen that the higher the sprocket speed, the larger the chain pitch, and the fewer the number of sprocket teeth, the dynamic load will increase

(2) the partial speed of the chain along the vertical direction also changes periodically, causing the chain to produce transverse vibration, which is also one of the reasons for the dynamic load of the chain drive

(3) when the chain link enters the sprocket, the chain link and the sprocket teeth mesh at a certain relative speed, and the sprocket and the teeth will be impacted and generate additional dynamic load. As shown in Figure 3, according to the principle of relative motion, the sprocket is regarded as stationary, and the chain link enters the gear teeth at an angular speed of -w to produce impact. This phenomenon intensifies with the increase of sprocket speed and chain pitch. Make the transmission produce vibration and noise

(4) if the chain tension is not good and the chain is loose, inertia impact will occur under the conditions of starting, braking, reverse rotation, load change, etc., causing great dynamic load to the chain drive

due to the dynamic load effect of chain drive, chain drive is not suitable for high speed

force analysis of chain drive

when installing chain drive, only a small tension is needed, mainly to make the sag of loose side of chain not too large, otherwise significant vibration, tooth skipping and chain disconnection will occur. If the dynamic load in the transmission is not considered, the forces acting on the chain include the circumferential force (i.e. effective tension) f, centrifugal tension FC and suspension tension FY found by Smithers Rapra company. As shown in the figure

the main forces of the chain in the transmission are:

(1) the tight side tension of the chain is f1=f+fc+fy (n)

(2) the loose side tension of the chain is f2=fc+fy (n)

(3) the chain link around the sprocket. During the movement, this investment will significantly increase the centrifugal tension generated for the real market supply guarantee ability of global customers fc=qv2 (n)

, where: q is the mass per meter of the chain, kg/m

v is the chain speed m/s

(4) suspension tension

fv=kvqga (n)

can be approximately obtained by calculating the suspension tension, where: A is the center distance of chain drive, m

g is the gravitational acceleration, g=9.81m/s2

kv is the sag coefficient when the sag y=0.02a, and the installation angle β For (Figure 12.12), see table

the pressure FQ of the chain acting on the shaft can be approximately taken as fq= (1.2~1.3) f, and the larger value is taken when there is impact and vibration

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