Answer:
t=2.025 inches
Explanation:
Given that
P = 400 Psi
Yield stress ,σ = 80 ksi
Diameter ,d= 45 ft
We know that
1 ft = 12 inches
d= 540 inches
Factor of safety ,K= 3
The required thickness given as
t=thickness
t=2.025 inches
Therefore thickness will be 2.025 inches.
Answer:
critical stress required for the propagation is 27.396615 × N/m²
Explanation:
given data
specific surface energy = 0.90 J/m²
modulus of elasticity E = 393 GPa = 393 × N/m²
internal crack length = 0.6 mm
to find out
critical stress required for the propagation
solution
we will apply here critical stress formula for propagation of internal crack
( σc ) = .....................1
here E is modulus of elasticity and γs is specific surface energy and a is half length of crack i.e 0.3 mm = 0.3 × m
so now put value in equation 1 we get
( σc ) =
( σc ) =
( σc ) = 27.396615 × N/m²
so critical stress required for the propagation is 27.396615 × N/m²