Calculate the maximum internal crack length allowable for a 7075-T651 aluminum alloy component that is loaded to a stress one-ha
1 answer:
Answer: 0.0033 m = 3.3 mm (0.13 in.)
Explanation:
since this problem requires us to calculate the maximum internal crack length allowable for the 7075-T651
aluminum alloy is given that it is loaded to a stress level equal to one-half of its yield strength.
For this alloy, KIc 24 MPa (22 ksi
); also, σ= σ_y/2 = (495 MPa)/2 = 248 MPa (36,000 psi).
Now solving for 2ac gives us;
2a_c=2/π (K_IC/γσ)^2=2/π [(24MPa √m)/(1.35)(248MPa) ]^2=0.0033m=3.3mm (0.13 in.)
You might be interested in
Answer:
The explanations are provided below.
Explanation:
a) The accelerated thermal cycling test is a test that is performed to evaluate the fatigue strength of electronic solder connections. The glass transition temperature given as is a frequently used parameter to test the degree of cure of an epoxy encapsulant.
b) The Coefficient of Thermal Expansion of Encapsulant
This is an important parameter to prevent crackling. Crackling occurs when a joint is subjcted to high temperatures during reflow soldering. Thus, the property is important in determining the limits of the joint.
Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness = 92 Mpa√m
yield strength σ = 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length = 1/π(
/ Yσ )²
we substitute
= 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
= 1/π( 92 Mpa√m / (517.5 Mpa )²
= 1/π( 0.177777 )²
= 1/π( 0.03160466 )
= 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection