Chapter 14

Q. 14.1

Finding the Spring Rate

A helical compression spring of an average coil diameter D , wire diameter d , and number of active coils N_a supports an axial load P (Figure 14.3a). Calculate the value of P that will cause a shear stress of \tau _{all} , the corresponding deflection, and rate of the spring.

Given: D =48 mm, d=6 mm, N_a =5, \tau _{all} =360 MPa

Design Decisions: A steel wire of G =79 GPa is used.



Verified Solution

The mean diameter of the spring is D =48 − 6=42 mm. The spring index is equal to C =42/6=7. Applying Equation (14.6), we have

\tau_t=K_s \frac{8 P D}{\pi d^3}=K_s \frac{8 P C}{\pi d^2}      (14.6)

360=\frac{8 P(42)}{\pi(6)^3}\left(1+\frac{0.615}{7}\right)=0.539 P


P = 668  N

From Equation (14.10),

\delta=\frac{8 P D^3 N_s}{G d^4}=\frac{8 P C^3 N_a}{G d}        (14.10)

\delta=\frac{8(668)(7)^3(5)}{\left(79 \times 10^3\right)(6)}=19.34  mm

The spring rate is therefore k =668/0.01934=34.54 kN/m.