In order to determine the relationship between the bending moment and the frictional moment, consider a small segment of the spring with length dl. Let the central angle corresponding to this micro-spring wire be dθ, and assume that the pressure on the left side of the segment is P, while the pressure on the right side is P + dP. The bending moment acting on both sides is M. Additionally, there is a tension force dF in the micro-segment, a normal force dN exerted by the housing onto the spring, and a friction coefficient f between the housing and the spring.
When the housing rotates in the same direction as the spring, the friction torque generated by the housing on the spring coil is Tf0. The friction force acting on the micro-segment from the housing is fdN. Under equilibrium conditions, we can write the horizontal and vertical force balance equations for the micro-segment:
$$
P \cos\left(\frac{d\theta}{2}\right) - fdN - (P + dP)\cos\left(\frac{d\theta}{2}\right) = 0 \quad \text{(1)}
$$
$$
dF + P \sin\left(\frac{d\theta}{2}\right) + (P + dP)\sin\left(\frac{d\theta}{2}\right) - dN = 0 \quad \text{(2)}
$$
From equation (1), we derive:
$$
dN = -\frac{dP}{f} \cos\left(\frac{d\theta}{2}\right) \quad \text{(3)}
$$
The negative sign indicates that the pressure between the housing and the spring decreases due to the overrunning motion.
Next, the relationship between the bending moment M and the tension dF in the spring wire is given by:
$$
dF = \frac{2M d\theta}{D} \quad \text{(4)}
$$
Substituting equations (3) and (4) into equation (2), and assuming small angles where $\sin\left(\frac{d\theta}{2}\right) \approx \frac{d\theta}{2}$ and $\cos\left(\frac{d\theta}{2}\right) \approx 1$, we simplify the equation to:
$$
\frac{dP}{2MD} + P = -f d\theta \quad \text{(5)}
$$
Now, let’s integrate this equation over an angular range from 0 to U, where the pressures at two cross-sections are P₠and P₂ respectively. After integration, we obtain:
$$
P_1 + \frac{2M}{D} = 2Me^{-fU} \quad \text{(7)}
$$
Since Pâ‚‚ = 0, the effective pull force is Pfâ‚€ = Pâ‚. Substituting into equation (7), we find:
$$
Pfâ‚€ D = 2M(e^{-fU} - 1)
$$
Therefore, the effective torque is:
$$
Tfâ‚€ = M(e^{-fU} - 1) \quad \text{(8)}
$$
If the number of contact turns between the spring and the housing is N, then $U = 2\pi N$. Substituting into equation (8), the friction torque during overrunning becomes:
$$
Tf = M(e^{-2\pi N f} - 1) \quad \text{(9)}
$$
Assuming the clutch transmits a known working torque T, the relationship between the bending moment M and the operating torque T is:
$$
M = T \cdot e^{2\pi N f} - 1 \quad \text{(10)}
$$
Substituting equation (10) into (9), we get:
$$
Tf = T \cdot \frac{e^{-2\pi N f} - 1}{e^{2\pi N f} - 1} \quad \text{(11)}
$$
Equation (11) shows that the overrunning friction torque is negative, indicating it acts in the opposite direction to the working torque. For practical purposes, we take its absolute value, resulting in:
$$
Tf = T \cdot \frac{1 - e^{-2\pi N f}}{e^{2\pi N f} - 1} \quad \text{(12)}
$$
This is the formula used to calculate the friction torque during overrunning in an expansion-type self-excited overrunning spring clutch. As shown in equation (12), the overrunning friction torque increases with higher operating torque and decreases with a higher friction coefficient.
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