Calculate the length of internal crack in a plate fabricated from a steel alloy that has a fracture of 86MPa m subjected to tensile stress of 360 MPa .
Answers
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).
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Thus the length of the internal crack is 18 mm.
Explanation:
Given data:
- Fracture = 86 M Pa
- Tensile stress = 360 M Pa
Solution:
λc = 1 / π [ k / y σc]^2
λc = 1 / 3.14 [ 85 Mpa / (1.0) (360) [^2
λc = 0.318 x (0.236)^2
λc = 0.318 x 0.0556
λc = 0.017 m
λc = 17 mm
Thus the length of the internal crack is 18 mm.