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Question 14.9: Automotive-Type Multileaf Spring: Design for Fatigue Loading......

Automotive-Type Multileaf Spring: Design for Fatigue Loading

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.

F14.17
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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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