A structural component is fabricated from an alloy that has a plane-strain fracture toughness of 62 MPa√m. It has been determined that this component fails at a stress of 250 MPa when the maximum length of an internal crack is 1.6 mm. What is the maximum allowable internal crack length (in mm) without fracture for this same component exposed to a stress of 250 MPa and made from another alloy that has a plane strain fracture toughness of 40 MPa√m?

0.67 mm

Explanation:

Solution:

We find the dimensionless parameters by applying the critical stress crack propagation formula stated below:

σс= Klc/Y√πa

Y = Klc/σс √πa

σс = this is the critical stress needed for initial cracking propagation

Klc = the plain stress fracture toughness

a = surface length of the crack

Y = the dimensionless parameter

Now, we substitute the values  62MPa√m for Klc, 250 MPa for σс and  1.6 * 10 ^⁻3 for a in the dimensionless parameter equation.

Thus,

Y = Klc/σс √πa

= 62/250(√π * 1.6* 10 ^⁻3)

= 3.492

The next step is to find the maximum permitted surface crack length by applying the  critical stress crack propagation equation given below:

σс= Klc/Y√πa

a= 1/π (Klc/Yσс)²

Now, substitute 40 MPa√m for Klc, 250 MPa for σс and 3.492 for surface length crack  equation

So,

a= 1/π (Klc/Yσс)²

= 1/π[40/3.492 * 250]²

=1/π[40/873]²

=1/π[0.0458]²

0.318[0.0458]²

=0.318[0.00209]

= 0.0066

0.67* 10 ^⁻3 m

= 0.67 mm

Therefore the maximum surface crack length produced is 0.67 mm

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explanation:

answer:

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