Athin plate of steel contains a crack of length 2a =16 mm, which is subjected to a stress of 350 mpa. the yield stress of the material is 1000 mpa. the radius of the plastic region is approximately

0.49 mm

Explanation:

Given:

Crack length, 2a = 16 mm

or

a = 8 mm = 0.008 m

Applied stress, σ = 350 MPa

Yield stress, = 1000 MPa

Now,

The stress intensity factor (K) is given as:

K = σ√(πa)

on substituting the values, we get

K = 350 × √(π × 0.008)

or

K = 55.48

also,

from Irwin's plastic zone correction factor

we have the formula

where,

is the radius of the plastic zone

on substituting the respective values, we get

or

0.00049 m

or

0.49 mm

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