In order to establish the relationship between the bending moment and the frictional moment, consider a small segment of the spring coil with length dl. Let the central angle corresponding to this micro-spring wire be dU, 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 sections is M, and the tension in the micro-segment is dF. Additionally, the outer casing exerts a normal force dN on the micro-segment, and the coefficient of friction between the casing and the spring is f.
When the outer casing rotates in the same direction as the spring, the friction torque exerted by the casing on the spring coil is Tf0. The frictional force acting on the micro-segment due to the casing is fdN. Under equilibrium conditions, we can derive the horizontal and vertical force balance equations for the micro-segment:
$$
P \cos\left(\frac{dU}{2}\right) - fdN - (P + dP)\cos\left(\frac{dU}{2}\right) = 0 \quad (1)
$$
$$
dF + P \sin\left(\frac{dU}{2}\right) + (P + dP)\sin\left(\frac{dU}{2}\right) - dN = 0 \quad (2)
$$
From equation (1), we obtain:
$$
dN = -\frac{dP}{f} \cos\left(\frac{dU}{2}\right) \quad (3)
$$
The negative sign indicates that the contact pressure between the housing and the spring decreases when the spring is over-rotated. The relationship between the bending moment M and the tension dF is given by:
$$
dF = \frac{2MdU}{D} \quad (4)
$$
where D is the diameter of the spring. Substituting equations (3) and (4) into equation (2), we get:
$$
\frac{2MdU}{D} + P \sin\left(\frac{dU}{2}\right) + (P + dP)\sin\left(\frac{dU}{2}\right) + dP \cos\left(\frac{dU}{2}\right)f = 0 \quad (5)
$$
Assuming that dU is very small, we approximate $\sin\left(\frac{dU}{2}\right) \approx \frac{dU}{2}$ and $\cos\left(\frac{dU}{2}\right) \approx 1$. Also, since $dP \cdot \sin\left(\frac{dU}{2}\right)$ is negligible, equation (5) simplifies to:
$$
\frac{dP}{2MD} + P = -f dU \quad (6)
$$
Now, let’s integrate this equation between two cross-sections of the spring wire where the pressures are P₠and P₂, and the angular difference between them is U. Integrating from U = 0 to U, and noting that at U = 0, P₂ = 0, we find:
$$
P_1 + \frac{2M}{D} = 2Me^{-fU} \quad (7)
$$
Since Pâ‚‚ = 0, the effective pull force is Pfâ‚€ = Pâ‚. Substituting into equation (7):
$$
Pf_0 D = 2M(e^{-fU} - 1) \quad (8)
$$
Therefore, the effective torque is:
$$
Tf_0 = M(e^{-fU} - 1) \quad (9)
$$
If the number of contact turns between the active and passive housing is N, then $U = 2\pi N$, and the overrunning friction torque becomes:
$$
Tf = M(e^{-2\pi N f} - 1) \quad (10)
$$
Given that the clutch transmits a known working torque T, the relationship between the bending moment M and T is:
$$
M = T \frac{e^{2\pi N f} - 1}{e^{2\pi N f}} \quad (11)
$$
Substituting equation (11) into equation (10), we get:
$$
Tf = T \frac{(1 - e^{-2\pi N f})}{e^{2\pi N f} - 1} \quad (12)
$$
This is the formula for calculating the overrunning friction torque in an expansion-type self-excited overrunning spring clutch. As shown in equation (12), the friction torque increases with the operating torque transmitted by the clutch and decreases with a higher friction coefficient. This result provides important insights for designing and analyzing such clutch systems.
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