Question

In one study the critical stress intensity factor for human bone was calculated to be 4.05 MN/m3/2. If the value of Y in Eq. (2.8) is 1.2 and there may be a 2 mm crack present in a bone specimen, what would be the maximum tensile stress that can be applied before fracture occurs?

Answer 1

To find the maximum tensile stress allowable before fracture in a human bone with a given fracture toughness, crack length, and geometric factor Y, you can use the formula σ = κIC / (√(π * a) * Y). Substituting the known values calculates the maximum tensile stress to prevent fractures.

Step-by-step explanation:

The question you've posed relates to the field of materials science and mechanics of materials, specifically addressing the fracture mechanics of human bone. In fracture mechanics, we use the concept of the critical stress intensity factor, also known as fracture toughness, to predict the stress at which a material with a crack will fracture.

The equation to calculate the maximum tensile stress (σ) before fracture occurs when a crack is present is given by:

σ = κIC / (√(π * a) * Y)

Where, κIC is the critical stress intensity factor, 'a' is the half-length of the internal crack, and Y is a dimensionless parameter related to the geometry of the crack and specimen. Given that κIC for human bone is 4.05 MN/m³/2, Y is 1.2, and the crack length (2a) is 2 mm, we can calculate σ as follows:

1. Convert the crack length to meters (2 mm = 0.002 m).
2. Calculate the maximum tensile stress using the above equation.

Substituting the given values into the equation, we get σ = 4.05 MN/m³/2 / (√(π * 0.001 m) * 1.2) = 4.05 MN/m³/2 / (√(0.00314159 * 0.001 m) * 1.2). After calculation, this results in the maximum tensile stress value allowable before the fracture occurs.

By following this approach, we can ensure that the bones are not subjected to stresses that could potentially lead to fractures, thereby safeguarding against injury.