Question

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa √m (50 ksi √in.). If, during service use, the plate is exposed to a tensile stress of 200 MPa (29,000 psi), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.

Answer 1

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

$\sigma_c$ = tensile stress = 200 Mpa

$K$ = plane strain fracture toughness= 55 Mpa$\sqrt{m}$

$\lambda$= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

$\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}}$   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for $\lambda$

Multiplying both sides of equation (1) by $Y\sqrt{\pi\lambda}$

$\sigma_c Y\sqrt{\pi\lambda}=K$   (2)

Sequaring both sides of equation (2):

$(\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2$  

$\sigma^2_c Y^2 \pi\lambda=K^2$   (3)

Dividing both sides by $\sigma^2_c Y^2 \pi$ we got:

$\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2$   (4)

Replacing the values into equation (4) we got:

$\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m$

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.