Respuesta :
Answer: 133.88 MPa approximately 134 MPa
Explanation:
Given
Plane strains fracture toughness, k = 26 MPa
Stress at which fracture occurs, σ = 112 MPa
Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m
Critical internal crack length, l' = 6 mm = 6*10^-3 m
We know that
σ = K/(Y.√πa), where
112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]
112 MPa = 26 MPa / Y.√(3.142 * 0.043)
112 = 26 / Y.√1.35*10^-2
112 = 26 / Y * 0.116
Y = 26 / 112 * 0.116
Y = 26 / 13
Y = 2
σ = K/(Y.√πa), using l'instead of l and, using Y as 2
σ = 26 / 2 * [√3.142 * (6*10^-3/2)]
σ = 26 / 2 * √(3.142 *3*10^-3)
σ = 26 / 2 * √0.009426
σ = 26 / 2 * 0.0971
σ = 26 / 0.1942
σ = 133.88 MPa
In this exercise we have to calculate the value of stress in terms of pressure, in this way we find:
133.88 MPa approximately 134 MPa
Given the information in the statement, we have:
- Plane strains fracture toughness is k = 26 MPa
- Stress at which fracture occurs is σ = 112 MPa
- Maximum internal crack length is l = 8.6 mm = 8.6*10^-3 m
- Critical internal crack length is l' = 6 mm = 6*10^-3 m
We know that the calculus the stress is:
[tex]\sigma = K/(Y\sqrt{\pi a} )\\112 MPa = 26 MPa / Y.\sqrt{3.142 * 8.6*10^{-3})/2}\\112 MPa = 26 MPa / Y\sqrt{3.142 * 0.043}\\112 = 26 / Y.\sqrt{1.35*10^{-2}}\\112 = 26 / Y * 0.116\\Y = 26 / 112 * 0.116\\Y = 26 / 13\\Y = 2[/tex]
So using the same formula but now replacing the value of Y by 2, we have:
[tex]\sigma = K/(Y.\sqrt{\pi a} )\\\sigma = 26 / 2 * [\sqrt{3.142 * (6*10^-3/2)}]\\\sigma = 26 / 2 * \sqrt{(3.142 *3*10^-3)}\\\sigma = 26 / 2 * \sqrt{0.009426}\\\sigma = 26 / 2 * 0.0971\\\sigma = 26 / 0.1942\\\sigma = 133.88 MPa[/tex]
See more about about stress level at brainly.com/question/1446338