A three-span continuous beam is to be selected to carry a uniformly distributed dead load of 4.7 kip/ft, including its self-weig
1 answer:
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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