ETBX Fatigue Crack Growth Module - Example 4.pdf
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Transcript of ETBX Fatigue Crack Growth Module - Example 4.pdf
ETBX Fatigue Crack Growth Module - Example 4
Help Contents | Module Library
Fatigue Crack Growth Rate An EngineersToolbox Calculation Module
Example Problems
Example 4
A wide plate with a semi-elliptical edge crack of is subject to an oscillatory stress as shown in Figure 1.
Figure 1. Semi-elliptical crack at the edge of a plate loaded in tension
The plate is manufactured from A36 steel. The yield strength σys is 36 ksi and the fracture toughness KIC = 50 ksi-in1/2. The crack growth rate is given by the Walker
equation:
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ETBX Fatigue Crack Growth Module - Example 4
where the material constants C, n, and m are given in terms of ksi and inches by:
Since the width b of the plate is assumed to be large compared to the crack length c, the following stress intensity factor solution is used:
where Q is a shape factor for an elliptical crack and is approximately given by:
Problem 1: For the constant amplitude load spectrum shown in Figure 2, calculate the number of cycles to failure given an initial crack length of ai = 0.10 inches. Assume the ratio of crack depth to crack length a/c = 0.50 and remains fixed during crack growth.
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ETBX Fatigue Crack Growth Module - Example 4
Figure 2. Load spectrum for Problem 1
Problem 2: Repeat Problem 1 for the load spectrum given in Figure 3.
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ETBX Fatigue Crack Growth Module - Example 4
Figure 2. Load spectrum for Problem 2
Solution
The first step in determining the number of cycles to failure is to calculate the critical crack length ac, which is given by:
The stress intensity correction factor β is a constant value due to the assumption that a/c remains constant, and due to the assumption that the plate width b is much greater than c. The β value is calculated by evaluating Q for a/c = 0.5 and substituting into the equation for KIC:
After solving for β , the ETBX Fatigue Crack Growth module can be used to calculate ac.
A picture of the module input form to calculate ac is shown in Figure 3 . Module output
is shown in Figure 4. The critical crack length for σmax = 23 ksi is calculated to be
1.758 inches.
The number of cycles to failure is calculated by choosing the Walker solution and entering ai = 0.10 inches and af = 1.758 inches. The material contants C, n, and m are based on units of ksi and inches, and thus all module inputs must be expressed in those units. The stress range ∆σ = (23 ksi - 13 ksi) = 10 ksi. The stress ratio R = (σmin /
σmax) = 13 ksi / 23 ksi = 0.565. The input form is shown in Figure 5.
Clicking the calculate button displays the solution in the output window shown in Figure 6 . The number of cycles to failure is computed to be 1,842,218 cycles.
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ETBX Fatigue Crack Growth Module - Example 4
The analysis can be easily repeated for the input spectrum given in Figure 2. The new spectrum requires that ac and R be recalculated. For σmax = 15 ksi and σmin = 5 ksi, ac
= 4.134 inches and R = 0.333. The number of cycles to failure is computed to be 2,640,201 cycles. An ETBX plot of crack growth rate da/dN versus stress intensity factor K for Problems 1 and 2 is shown in Figure 7. The crack growth plots are compared in Figure 8.
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ETBX Fatigue Crack Growth Module - Example 4
Figure 3. Module input form for calculation of critical crack length
Figure 4. Module output for critical crack length calculation
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ETBX Fatigue Crack Growth Module - Example 4
Figure 5. Module input form for cycles to failure calculation
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ETBX Fatigue Crack Growth Module - Example 4
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ETBX Fatigue Crack Growth Module - Example 4
Figure 6. Module output for cycles to failure calculation
Figure 7. Crack growth rate curves for R = 0.565 and R = 0.333
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ETBX Fatigue Crack Growth Module - Example 4
Figure 8. Comparison of crack growth plots
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ETBX Fatigue Crack Growth Module - Example 4
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