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10/31/13

Fatigue Solution 3

FATIGUE CRACK GROWTH

PROBLEM 3

SOLUTION

This problem illustrates the case where a cracked structure will not meet the fatigue life requirement, and explores the possibilities of increasing the cyclic life. It is straightforward and should take around 20 minutes to complete.

_______________________________________ A structure contains a critical component made from A514 steel. After fabrication of the structure, a welding defect 7.6 mm deep is discovered in this steel plate. The flaw is essentially an edge crack under tension loading, and the required cyclic life of the structure is 100 000 cycles. The component is subject to a fluctuating load which causes a stress variation from 172 MPa to 310 MPa. Material properties for the A514 steel are: yield stress = 689 MPa, K1C = 165 MPa m½ geometry correction factor Y = 1.12, and the Paris law is:

where da/dN is in m/cycle, and:

i) Calculate the fatigue life of this component based on attaining a critical defect size for fast fracture. ii) Accurately construct the curve showing crack length against number of applied load cycles. iii) Discuss the various measures that could be adopted to extend the life of the structure to 100 000 cycles. iv) What is the effect of reducing the initial defect size to 5 mm (by weld repair with better control of process parameters)? Explain this result in terms of the shape of the curve of crack length versus cyclic life. Scientific Calculator

RPN Calculator Theory

_______________________________________ Answer: Nf = 87 992 cycles

_______________________________________ i) In order to calculate the cyclic life, we require the critical crack size causing fast fracture. This can be obtained by substituting the appropriate data into the K calibration equation:

i.e. 71.9 mm. Hence the limits on the Paris law integration are 7.6 mm and 71.9 mm, while the stress range is (310 - 172) MPa = 138 MPa. Separating the variables and rearranging www.tech.plym.ac.uk/sme/tutorials/FMTut/Fatigue/Solutions/Solution3.htm

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Fatigue Solution 3

the Paris law gives:

We can perform the integration symbolically and then substitute the actual values into the equation to obtain the required life.

ii) To draw the accurate crack length - life curve, all that is required is to calculate N at various lengths between 7.6 mm and 71.9 mm, i.e. repeating the above calculation with various af values. The table below gives some typical values, and the curve is drawn in the figure. a (m)

0.015

0.025

0.04

0.06

N

29 421

49 850

67 486

81 857

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iii) Fatigue life can be extended by a number of measures, all of which rely on changing relevant parameters in the Paris law. These are stress range, final crack length and initial crack length. Stress range can be reduced by decreasing the applied loads, which is usually not feasible particularly for new structures, or by increasing component size (which may also not be feasible). Sometimes, when a crack is detected in service, e.g. in a pressure vessel, it may be possible to reduce applied stresses and continue operation until a scheduled maintenance period. Final crack length could be increased if it was possible to increase the toughness locally, e.g. by using additional material at the crack location, or by replacing the cracked component with one of a higher toughness (and hence a larger critical defect size for the same value of applied peak stress. The increase in life would, however, be quite low because of the fast growth rates as the crack approaches the critical size. The best solution is to reduce the initial defect size, either by repairing or replacing the component. If the initial defect size is likely to be an intrinsic problem in the fabrication/repair process for this component, the fabrication process must be controlled more closely to reduce the initial crack length. iv) Reducing the initial defect to 5 mm gives a cyclic life of 107 423 cycles. The figure above indicates that the crack growth rate curve is exponential and hence a relatively small reduction in initial crack length yields a large life return. Next Problem

Introduction

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