thể 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 = 40x 10-15 (AK) mm/cycle 38920 cycles O 26919 cycles O 38720 cycles O 17929 cycles

Materials Science And Engineering Properties
1st Edition
ISBN:9781111988609
Author:Charles Gilmore
Publisher:Charles Gilmore
Chapter11: Fracture And Fatigue
Section: Chapter Questions
Problem 7ETSQ
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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 = 40x10 15 (AK)'
mm/cycle
38920 cycles
O 26919 cycles
38720 cycles
17929 cycles
Transcribed Image Text: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 = 40x10 15 (AK)' mm/cycle 38920 cycles O 26919 cycles 38720 cycles 17929 cycles
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