Answer:
the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Explanation:
Given the data in the question;
yield strength σ
= 690 Mpa
plane strain fracture toughness K
= 32 MPa-
minimum component thickness for which the condition of plane strain is valid = ?
Now, for plane strain conditions, the minimum thickness required is expressed as;
t ≥ 2.5( K
/ σ
)²
so we substitute our values into the formula
t ≥ 2.5( 32 / 690 )²
t ≥ 2.5( 0.0463768 )²
t ≥ 2.5 × 0.0021508
t ≥ 0.005377 m or 5.38 mm
Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Answer:
Cite two important additional refinements that resulted from the Wave-mechanical
atomic model.
Answer:
Magnitude of force P needed to start
towing the 40-kg crate = 140.14 N
Location of the resultant normal force acting on the crate, measured from point A = 500
Explanation:
The explanation for this question is given in the attachment below.