**Stability of a Hard-Drawn Wire Compression Spring**

Reconsider the hard-drawn compression spring discussed in Example 14.3.

**Find:**

a. The solid length.

b. Whether the spring will buckle in service.

c. The pitch of the body coil.

**Assumptions:** The modulus of rigidity of the wire will be G =79 GPa. The spring rate equals k =1.4 kN/m.

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Refer to the numerical values given in Example 14.3. The solid deflection is:

\delta_s=\frac{P_{ all }}{k}=\frac{45}{100}=0.03214 m =32.14 mm

a. The number of active coils, by Equation (14.11), is

k=\frac{P}{\delta}=\frac{G d^4}{8 D^3 N_s}=\frac{d G}{8 C^3 N_a} (14.11)

N_a=\frac{G d}{8 k C^3}=\frac{\left(79 \times 10^9\right)\left(1.88 \times 10^{-3}\right)}{8(1400)\left(9^3\right)}=18.19

For the squared and ground ends, observe from Figure 14.7(d) that

N_t=N_a+20.19

and the solid length

h_s=N_t d=(20.19)(1.88)=37.96 mm

b. Applying Equation (14.14), the free length is equal to

\delta_s=h_f-h_s

h_t=h_s+\delta_s=37.96+32.14=70.1 mm

For the case under consideration, we have

\frac{\delta_s}{h_f}=\frac{32.14}{70.1}=0.46 \quad \text { and } \quad \frac{h_f}{D}=\frac{70.1}{16.92}=4.14

Case A in Figure 14.10 illustrates that the spring is far outside of the buckling zone and obviously safe.

c. From Figure 14.7(d), the pitch is

p=\frac{1}{N_a}\left(h_f-2 d\right)=\frac{1}{18.19}[70.1-2(1.88)]=3.647 mm

Comment: With the values of D , N_t , and h_f obtained here and in the previous example, a compression spring can be drawn or made.