Question

A structural component in the shape of a flat plate 25.0 mm thick is to be fabricated from a metal alloy for which the yield strength and plane strain fracture toughness values are 545 MPa and 29.6 MPa-m1/2, respectively. For this particular geometry, the value of Y is 1.3. Assuming a design stress of 0.3 times the yield strength, calculate the critical length of a surface flaw.

Answer 1

Answer:

The critical length of a surface flaw is 6.17 mm

Explanation:

according to the exercise:

flat plate size=25 mm

σ=545 MPa

plane strain fracture toughness=29.6 MPa-m1/2

Y=1.3

Design stress = 0.3*σ = 163.5 MPa

For plain structure fracture toughness is:

$K_{IC} =Y*o\sqrt{\pi a_{c} }$

Where ac is the maximum allowable flat size. Clearing ac:

$a_{c} =\frac{1}{\pi } (\frac{K_{IC} }{o*Y} )^{2} =\frac{1}{\pi } (\frac{29.6}{163.5*1.3} )^{2} =6.17x10^{-3}m=6.17mm$

Answer 2

Answer:

The critical length of surface flaw = 6.176 mm

Explanation:

Given data-

Plane strain fracture toughness Kc = 29.6 MPa-m1/2

Yield Strength = 545 MPa

Design stress. =0.3 × yield strength

= 0.3 × 545

= 163.5 MPa

Dimensionless parameter. Y = 1.3

The critical length of surface flaw is given by

= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2

Now putting values in above equation we get,

= 1/3.14( 29.6 / 1.3 × 163.5)^2

=6.176 × 10^-3 m

=6.176 mm