A six-leaf spring is subjected to a load at the center that varies between $P_{max}$ and $P_{min}$ (Figure 14.17). Estimate the total length $2L$ and width of each leaf.

Given: $P_{min}$=160 lb, $P_{max}$=800 lb, $n$=6.

Assumptions: Stress concentration at the center is such that $K_f$=1.2. Use a survival rate of 50% and $C_f=C_e$=1.

Design Decisions: We use a steel alloy spring of $S_u$ = 200 ksi, $S_e^{\prime}=78 ksi , E=30 \times 10^6 psi , \nu=0.3$, $h$ = 0.25 in., and $k$ = 140 lb/in. The material is shot peened. A safety factor of ns = 1.4 is applied.

Step-by-Step
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From Table 7.3, $C_r$=1. The modified endurance limit, by Equation (7.6), $S_e$=(1)(1)(1)(1/1.2)78=65 ksi.

$S_e=C_f C_r C_s C_t\left(1 / K_f\right) S_e^{\prime}$     (7.6)

Each half of a spring acts as a cantilever supporting half of the total load. The mean and the alternating loads are therefore

$P_m=\frac{400+80}{2}=240 lb , \quad P_a=\frac{400-80}{2}=160 lb$

Inasmuch as bending stress is directly proportional to the load, we have $\sigma_a / \sigma_m=P_a / P_m=2 / 3 .$

The mean stress, using Equation (14.42), is

$\sigma=\frac{6 P L}{n b h^2}$     (14.42)

$\sigma_m=\frac{6 P_m L}{n b h^2}=\frac{6(240) L}{6 b(0.25)^2}=3840 \frac{L}{b}$     (e)

Substituting the given numerical values into Equation (7.20), we have

$\sigma_m=\frac{S_u / n}{\frac{\sigma_a}{\sigma_m} \frac{S_u}{S_e}+1}$       (7.20)

$\sigma_m=\frac{S_u / n_s}{\frac{\sigma_a}{\sigma_m} \frac{S_u}{S_e}+1}=\frac{200 / 1.4}{\frac{2}{3} \frac{200}{65}+1}=46.82 ksi$

From Equation (e),

$48,820=3840 \frac{L}{b} \quad \text { or } \quad b=0.082 L$     (f)

Because the spring is loaded at the center with $2P$, Equation (14.44) becomes $k=E n b h^3 / 3 L^3\left(1- \nu^2\right)$. Introducing the given data results in

$k=\frac{P}{\delta} \frac{E n b}{6\left(1- \nu^2\right)}\left(\frac{h}{L}\right)^3$     (14.44)

$140=\frac{\left(30 \times 10^6\right)(6)(0.82 L)(0.25)^3}{3 L^3(0.91)}$

$L = 24.56 in.$

Hence, the overall length is $2L$=49.12 in. The width of each of the six leaves using Equation (f) equals b=0.082(24.56)=2.014 in.

 TABLE 7.3 Reliability Factors Survival Rate (%) $C_r$ 50 1.00 90 0.89 95 0.87 98 0.84 99 0.81 99.9 0.75 99.99 0.70

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