Question

What is the magnitude of the maximum stress that exist at the tip of an internal crack having a radius of curvature of 1.9 x 10-4 mm and a crack length of 3.8 x 10-2 mm when a tensile stress of 140 MPa is applied?

Answer 1

Answer:

2800 [MPa]

Explanation:

In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

$\sigma_{m} = 2 \sigma_{p} (\frac{l_{c}}{r_{c}})^{0.5}$

Where $\sigma_{m}$ is the magnitude of he maximum stress at the tip of the crack, $\sigma_{p}$ is the magnitude of the tensile stress, $l_{c}$ is $1/2$ the length of the internal crack, and $r_{c}$ is the radius of curvature of the crack.

We have:

$r_{c}=1.9*10^{-4} [mm]$

$l_{c}=3.8*10^{-2} [mm]$

$\sigma_{c}=140 [MPa]$

We replace:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}$

We get:

$\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]$

Answer 2

The magnitude of the maximum stress that exist at the tip of an internal crack is equal to 280 Mpa.

Given the following data:

Radius of curvature = 1.9 x $10^{-4}$ mm.

Crack length = 3.8 x $10^{-2}$ mm.

Tensile stress = 140 MPa.

How to calculate the maximum stress.

Mathematically, the magnitude of the maximum stress that exist at the tip of an internal crack can be calculated by using this formula:

$\sigma_m = 2\sigma_p( \frac{L}{r} )^\frac{1}{2}$

Where:

• $\sigma_p$ is the tensile stress.

• L is the length of internal crack.

• r is the radius of curvature.

Note: The total length would be divided by 2.

Substituting the given parameters into the formula, we have;

$\sigma_m = 2\times 140 \times ( \frac{1.9 \times 10^{-2}}{1.9 \times 10^{-2} } )^\frac{1}{2}\\\\\sigma_m = 2\times 140 \times 1$

Maximum stress = 280 Mpa.

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