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

0.024 m = 24.07 mm

Step-by-step explanation:

1) Notation

`\sigma_c` = tensile stress = 200 Mpa

`K` = plane strain fracture toughness= 55 Mpa
`√(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=(K)/(Y√(\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√(\pi\lambda)`

`\sigma_c Y√(\pi\lambda)=K` (2)

Sequaring both sides of equation (2):

`(\sigma_c Y√(\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=(1)/(\pi)[(K)/(Y\sigma_c)]^2` (4)

Replacing the values into equation (4) we got:

`\lambda=(1)/(\pi)[(55 Mpa√(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.