Lug Fatigue Analysis

Lug Fatigue Analysis Nomenclature

9.17 Lug Fatigue Analysis

A method for determining the fatigue strength of 2024-T3 and 7075-T6 aluminum alloy lugs under axial loading is presented.

Figures 9-21 and 9-22 show the lug and the range of lug geometries covered by the fatigue strength prediction method. Fatigue lives for lugs having dimensional ratios falling outside the region shown should be corroborated by tests.

In this method the important fatigue parameters are k₁, k₂, and k₃ (see Figure 9-23).

To find the allowable life knowing the applied stresses and lug dimensions, or to find the allowable stresses knowing the life, R value (R = f_min/f_max) and lug dimensions, use the following procedure:

1. Enter Figure 9-22 to check that the lug dimensional ratios fall within the region covered by the method. Enter Figure 9-23 and read k₁, k₂, and k₃; calculate the product k₁k₂k₃.
2. Calculate the allowable net-tension static stress for the lug, F_nu.L, according to the method described in Section 9.3.2.
3. Determine the value 0.4F_nu.L. This is the alternating stress corresponding to a maximum stress value of 0.8F_nu.L when f_min = 0.8F_nu.L was chosen as an average yield stress value for 2024 and 7075 aluminum alloy lugs.
4. Using the value 0.4F_nu.L as an alternating stress, draw a straight line between the intersection of this value and the appropriate k₁k₂k₃ curve on Figures 9-24 or 9-25, and the point 0.5F_nu.L at 1 cycle. This extends the k₁k₂k₃ curve to cover the entire life range to static failure.
5. Enter Figure 9-24 or 9-25 (lug fatigue curves for the case where R = 0) with k₁k₂k₃. For values of life, N = 10³, 3×10³, 10⁴, etc., or any other convenient values, determine the corresponding values of f_a, the stress amplitude causing fatigue failure when R = 0.
6. Plot the values of f_a found in Step 5 along the R = 0 line in a Goodman diagram such as shown in Figure 9-26 (f_m = f_a when R = 0). The Goodman diagram shown in Figure 9-27 applies to a particular 7075-T6 lug for which k₁k₂k₃ = 1.32 (see example problem 1), but is typical of all such diagrams.
7. Plot the allowable net-tension static stress found in Step 2 as f_m at the point (f_m, 0) of the Goodman diagram (f_m = f_max when f_a = 0). For the case considered in Figure 9-26, this point is plotted as (f_m = 70,000 psi, f_a = 0).
8. Connect the point plotted in Step 7 with each of the points plotted in Step 6 by straight lines. These are the constant life lines for the particular lug being analyzed. The Goodman diagram is now complete and may be used to determine a life for any given applied stresses, or to determine allowable stresses knowing the life and R value.

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9.18 Example Problem of Lug Fatigue Analysis

Given a concentric 7075-T6 aluminum lug as shown in Figure 9-21, with the following dimensions: a = 0.344 in, c = 0.3444 in, and D = 0.437 in. If the lug is subjected to a cycle axial load such that the maximum net-section stress is 27,000 psi and the minimum net section stress is 18,470 psi, find the fatigue life.

From the lug dimensions,

a/c = 1.0;  c/D = 0.787;  (D/c = 1.27)

(1) Figure 9-22 indicates that the lug may be analyzed using this method. From Figure 9-23,

k₁ = 1.0;  k₂ = 1.33;  k₃ = 0.99; k₁k₂k₃ = 1.32

(2) Calculate the allowable net-section tensile ultimate stress, F_nu.L, for Equation (9-4) in Section 9.3.2. For the given lug, F_nu.L = 70,000 psi.

(3) 0.4F_nu.L = 0.4 × 70,000 = 28,000 psi.

(4) Draw a light pencil line on Figure 9-24 from the point (f_a = 28,000 psi on k₁k₂k₃ = 1.32) to the point (f_a = 35,000, N = 1 cycle) (This is illustrated, for clarity, on Figure 9-27).

(5) Enter Figure 9-24 and read values of f_a for various numbers of life cycles, using the line k₁k₂k₃ = 1.32. These numbers are as follows:

N	102	103	3×103	104	3×104	105	106	107
fa	30 KSI	24.5	18.8	13.5	8.88	5.70	2.34	1.30

(6) Plot the values of f_a along the R = 0 line of the Goodman diagram. (Refer to Figure 9-26.)

(7) Plot F_nu.L = 70,000 psi, as f_m at the point (f_m, 0) of the Goodman diagram. (Refer to Figure 9-26.)

(8) Connect the points plotted in Step 6 with the point plotted in Step 7 by straight lines. The Goodman diagram is now complete.

(9) Enter the Goodman diagram with values of \( f_a = {27,700 - 18,470 \over 2} = 4,615 ~\text{psi} \) and \( f_m = {27,700 + 18,470 \over 2} = 23,085 ~\text{psi} \), and read the fatigue life, N = 8×10⁴ cycles, by interpolation (test results show N = 8.6×10⁴ cycles).

If the known quantities are life and R values, e.g., N = 10⁴ cycles and R = 0, the allowable stresses can be obtained by using the same Goodman diagram. Enter the completed Goodman diagram at R = 0 and N = 10⁴ cycles and read the amplitude and mean stresses (in this case f_a = f_m = 13,500 psi).

Only if the lug dimensions are changed, must a new Goodman diagram be drawn.

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