Question

calculate the length of the internal crack in a plate fabricated

Answer 1

Answer:

HELLO DEAR!

Explanation:

A big plate is manufactured from an iron alloy that has a flat strain rupture thickness of 82.4MPam−−√(75.0ksiin.−−√). If the plate is shown to a tensile strain of 345 MPa (50,000 psi) through service use, define the smallest length of an exterior crack that will guide to fracture. Assume a rate of 1.0 for Y.

For an exterior crack;

a=0.00001 m∧f =1.12

So σ=KIC/ f√πa=45 MPa√m/(1.12 ×√π∗0.00001 m)=¿ 7168.3 MPa ¿

Since the used strain needed for breakdown due to fracture distribution is still essential than 550 MPa, the ceramic is supposed to fail due to overload and not because of the defects (related to the physical cracks).

PLEASE MARK AS BRAINLIEST!!!!

Answer 2

Answer:A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4MPam−−√82.4MPa  

m

​  

(75.0ksiin.−−√)(75.0ksi  )

in.

​  

. If the plate is exposed to a tensile stress of 345 MPa (50,000 psi) during service use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.

Explanation: i hope you got it