Calculate the count of cycles resulting in breakdown."Given : Crack length= 0.2 mm3Fracture toughness= 1708 N/mm 2Cyclic loading= 175 N/mm2Crack growth rate = 40x10 15 (AK)'mm/cycle38920 cyclesO 26919 cycles38720 cycles17929 cycles

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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

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

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Crack Detection

Concrete materials are meant for supporting structures, building foundations and they act as several load-bearing structures. Over time, due to environmental conditions, and based on the constituent materials used, concrete materials are subjected to a great degree of distress. Corrosion of the reinforcement bars due to chemical influences, uneven base, earthquake, drying shrinkages, and alkali-aggregate reactions are some of the reasons for concrete distress.

Question

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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