## 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)

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

Solving

$P = 668 N$

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.