Answer:
critical stress = 595 MPa
Explanation:
given data
fracture toughness = 74.6 MPa-
crack length = 10 mm
f = 1
solution
we know crack length = 10 mm
and crack length = 2a as given in figure attach
so 2a = 10
a = 5 mm
and now we get here with the help of plane strain condition , critical stress is express as
critical stress = ......................1
put here value and we get
critical stress =
critical stress = 595 MPa
so here stress is change by plane strain condition because when plate become thinner than condition change by plane strain to plain stress.
plain stress condition occur in thin body where stress through thickness not vary by the thinner section.
Answer:
option B is correct. Fracture will definitely not occur
Explanation:
The formula for fracture toughness is given by;
K_ic = σY√πa
Where,
σ is the applied stress
Y is the dimensionless parameter
a is the crack length.
Let's make σ the subject
So,
σ = [K_ic/Y√πa]
Plugging in the relevant values;
σ = [50/(1.1√π*(0.5 x 10^(-3))]
σ = 1147 MPa
Thus, the material can withstand a stress of 1147 MPa
So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.
Answer:
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Explanation:
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