Calculate the count of cycles resulting in breakdown. Given Crack length = 0.2 mm 3 Fracture toughness = 1708 N/mm ² Cyclic loading = 175 N/mm² Crack growth rate = 40×10-15 (AK)4 mm/cycle

Materials Science And Engineering Properties
1st Edition
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Charles Gilmore
Chapter11: Fracture And Fatigue
Section: Chapter Questions
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Calculate the count of cycles resulting in breakdown.
Given Crack length = 0.2 mm
3
Fracture toughness = 1708 N/mm²
Cyclic loading = 175 N/mm²
Crack growth rate = 40x10-15 (AK)4 mm/cycle
17929 cycles
38720 cycles
38920 cycles
26919 cycles
Transcribed Image Text:Calculate the count of cycles resulting in breakdown. Given Crack length = 0.2 mm 3 Fracture toughness = 1708 N/mm² Cyclic loading = 175 N/mm² Crack growth rate = 40x10-15 (AK)4 mm/cycle 17929 cycles 38720 cycles 38920 cycles 26919 cycles
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